Circle line-segment collision detection algorithm

Question

I have a line from A to B and a circle positioned at C with the radius R      What is a good algorithm to use to check whether the line intersects the circle  And at what coordinate along the circles edge it occurred

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No one seems to consider projection  am I completely off track here   Project the vector AC onto AB  The projected vector  AD  gives the new point D  If the distance between D and C is smaller than  or equal to  R we have an intersection   Like this

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Another solution  first considering the case where you don t care about collision location  Note that this particular function is built assuming vector input for xB and yB but can easily be modified if that is not the case  Variable names are defined at the start of the function  Line segment points  A0  Af  defined by xA0  yA0  xAf  yAf  circle center denoted by xB  yB  rB radius of circle  rA   radius of point  set to zero for your application  def staticCollision f xA0  yA0  xAf  yAf  rA  xB  yB  rB    note potential speed up here by casting all variables to same type and or using Cython           Build equations of a line for linear agents  convert y   mx   b to ax   by   c   0 means that a   -m  b   1  c   -b     m v    yAf - yA0     xAf - xA0      b v   yAf - m v   xAf     rEff   rA   rB  radii are added since we are considering the agent path as a thin line       Check if points  circles  are within line segment  find center of line segment and check if circle is within radius of this point      segmentMask   np sqrt   yB -  yA0 yAf  2   2    xB -  xA0 xAf  2   2    lt  np sqrt   yAf - yA0   2    xAf - xA0   2     2   rEff       Calculate perpendicular distance between line and a point     dist v   np abs -m v   xB   yB - b v    np sqrt m v  2   1      collisionMask    dist v  lt  rEff   amp  segmentMask       return True if collision is detected     return collisionMask  collisionMask any    If you need the location of the collision  you can use the approach detailed on this site  and set the velocity of one of the agents to zero  This approach works well with vector inputs as well  http   twobitcoder blogspot com 2010 04 circle-collision-detection html

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I would use the algorithm to compute the distance between a point  circle center  and a line  line AB   This can then be used to determine the intersection points of the line with the circle   Let say we have the points A  B  C  Ax and Ay are the x and y components of the A points  Same for B and C  The scalar R is the circle radius   This algorithm requires that A  B and C are distinct points and that R  is not 0    Here is the algorithm     compute the euclidean distance between A and B LAB   sqrt   Bx-Ax     By-Ay          compute the direction vector D from A to B Dx    Bx-Ax  LAB Dy    By-Ay  LAB     the equation of the line AB is x   Dx t   Ax  y   Dy t   Ay with 0  lt   t  lt   LAB      compute the distance between the points A and E  where    E is the point of AB closest the circle center  Cx  Cy  t   Dx  Cx-Ax    Dy  Cy-Ay          compute the coordinates of the point E Ex   t Dx Ax Ey   t Dy Ay     compute the euclidean distance between E and C LEC   sqrt  Ex-Cx     Ey-Cy         test if the line intersects the circle if  LEC  lt  R            compute distance from t to circle intersection point     dt   sqrt  R   - LEC            compute first intersection point     Fx    t-dt  Dx   Ax     Fy    t-dt  Dy   Ay         compute second intersection point     Gx    t dt  Dx   Ax     Gy    t dt  Dy   Ay       else test if the line is tangent to circle else if  LEC    R          tangent point to circle is E  else        line doesn t touch circle

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Here s an implementation in Javascript  My approach is to first convert the line segment into an infinite line then find the intersection point s   From there I check if the point s  found are on the line segment  The code is well documented  you should be able to follow along    You can try out the code here on this live demo  The code was taken from my algorithms repo        Small epsilon value var EPS   0 0000001      point  x  y  function Point x  y      this x   x    this y   y        Circle with center at  x y  and radius r function Circle x  y  r      this x   x    this y   y    this r   r        A line segment  x1  y1    x2  y2  function LineSegment x1  y1  x2  y2      var d   Math sqrt   x1-x2   x1-x2     y1-y2   y1-y2       if  d  lt  EPS  throw  A point is not a line segment     this x1   x1  this y1   y1    this x2   x2  this y2   y2        An infinite line defined as  ax   by   c function Line a  b  c      this a   a  this b   b  this c   c       Normalize line for good measure   if  Math abs b   lt  EPS        c    a  a   1  b   0      else        a    Math abs a   lt  EPS    0   a   b      c    b  b   1             Given a line in standard form  ax   by   c and a circle with     a center at  x y  with radius r this method finds the intersection    of the line and the circle  if any    function circleLineIntersection circle  line       var a   line a  b   line b  c   line c    var x   circle x  y   circle y  r   circle r        Solve for the variable x with the formulas  ax   by   c  equation of line       and  x-X  2    y-Y  2   r 2  equation of circle where X Y are known  and expand to obtain quadratic        a 2   b 2 x 2    2abY - 2ac   - 2b 2X x    b 2X 2   b 2Y 2 - 2bcY   c 2 - b 2r 2    0      Then use quadratic formula X    -b  - sqrt a 2 - 4ac   2a to find the       roots of the equation  if they exist  and this will tell us the intersection points       In general a quadratic is written as  Ax 2   Bx   C   0       a 2   b 2 x 2    2abY - 2ac   - 2b 2X x    b 2X 2   b 2Y 2 - 2bcY   c 2 - b 2r 2    0   var A   a a   b b    var B   2 a b y - 2 a c - 2 b b x    var C   b b x x   b b y y - 2 b c y   c c - b b r r        Use quadratic formula x    -b  - sqrt a 2 - 4ac   2a to find the       roots of the equation  if they exist      var D   B B - 4 A C    var x1 y1 x2 y2        Handle vertical line case with b   0   if  Math abs b   lt  EPS            Line equation is ax   by   c  but b   0  so x   c a     x1   c a          No intersection     if  Math abs x-x1   gt  r  return             Vertical line is tangent to circle     if  Math abs  x1-r -x   lt  EPS    Math abs  x1 r -x   lt  EPS        return  new Point x1  y         var dx   Math abs x1 - x       var dy   Math sqrt r r-dx dx           Vertical line cuts through circle     return         new Point x1 y dy         new Point x1 y-dy               Line is tangent to circle     else if  Math abs D   lt  EPS         x1   -B  2 A       y1    c - a x1  b       return  new Point x1 y1          No intersection     else if  D  lt  0         return          else        D   Math sqrt D        x1    -B D   2 A       y1    c - a x1  b       x2    -B-D   2 A       y2    c - a x2  b       return         new Point x1  y1         new Point x2  y2                     Converts a line segment to a line in general form function segmentToGeneralForm x1 y1 x2 y2      var a   y1 - y2    var b   x2 - x1    var c   x2 y1 - x1 y2    return new Line a b c         Checks if a point  pt  is inside the rect defined by  x1 y1    x2 y2  function pointInRectangle pt x1 y1 x2 y2      var x   Math min x1 x2   X   Math max x1 x2     var y   Math min y1 y2   Y   Math max y1 y2     return x - EPS  lt   pt x  amp  amp  pt x  lt   X   EPS  amp  amp           y - EPS  lt   pt y  amp  amp  pt y  lt   Y   EPS        Finds the intersection s  of a line segment and a circle function lineSegmentCircleIntersection segment  circle       var x1   segment x1  y1   segment y1  x2   segment x2  y2   segment y2    var line   segmentToGeneralForm x1 y1 x2 y2     var pts   circleLineIntersection circle  line         No intersection   if  pts length     0  return        var pt1   pts 0     var includePt1   pointInRectangle pt1 x1 y1 x2 y2         Check for unique intersection   if  pts length     1        if  includePt1  return  pt1       return            var pt2   pts 1     var includePt2   pointInRectangle pt2 x1 y1 x2 y2         Check for remaining intersections   if  includePt1  amp  amp  includePt2  return  pt1  pt2     if  includePt1  return  pt1     if  includePt2  return  pt2     return

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You can find a point on a infinite line that is nearest to circle center by projecting vector AC onto vector AB  Calculate the distance between that point and circle center  If it is greater that R  there is no intersection  If the distance is equal to R  line is a tangent of the circle and the point nearest to circle center is actually the intersection point  If distance less that R  then there are 2 intersection points  They lie at the same distance from the point nearest to circle center  That distance can easily be calculated using Pythagorean theorem  Here s algorithm in pseudocode     dX   bX - aX  dY   bY - aY  if   dX    0   amp  amp   dY    0            A and B are the same points  no way to calculate intersection   return       dl    dX   dX   dY   dY   t     cX - aX    dX    cY - aY    dY    dl      point on a line nearest to circle center nearestX   aX   t   dX  nearestY   aY   t   dY   dist   point dist nearestX  nearestY  cX  cY    if  dist    R           line segment touches circle  one intersection point   iX   nearestX    iY   nearestY     if  t  lt  0    t  gt  1               intersection point is not actually within line segment           else if  dist  lt  R           two possible intersection points    dt   sqrt R   R - dist   dist    sqrt dl         intersection point nearest to A   t1   t - dt    i1X   aX   t1   dX    i1Y   aY   t1   dY    if  t1  lt  0    t1  gt  1               intersection point is not actually within line segment             intersection point farthest from A   t2   t   dt    i2X   aX   t2   dX    i2Y   aY   t2   dY    if  t2  lt  0    t2  gt  1               intersection point is not actually within line segment           else          no intersection         EDIT  added code to check whether found intersection points actually are within line segment

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Weirdly I can answer but not comment    I liked Multitaskpro s approach of shifting everything to make the centre of the circle fall on the origin  Unfortunately there are two problems in his code  First in the under-the-square-root part you need to remove the double power  So not   var underRadical   Math pow  Math pow r 2   Math pow m 2  1   2 -Math pow b 2     but   var underRadical   Math pow r 2   Math pow m 2  1   - Math pow b 2    In the final coordinates he forgets to shift the solution back  So not   var i1    x t1 y m t1 b   but   var i1    x t1 c x  y m t1 b c y    The whole function then becomes   function interceptOnCircle p1  p2  c  r          p1 is the first line point       p2 is the second line point       c is the circle s center       r is the circle s radius      var p3    x p1 x - c x  y p1 y - c y     shifted line points     var p4    x p2 x - c x  y p2 y - c y        var m    p4 y - p3 y     p4 x - p3 x     slope of the line     var b   p3 y - m   p3 x    y-intercept of line      var underRadical   Math pow r 2  Math pow m 2    Math pow r 2  - Math pow b 2     the value under the square root sign       if  underRadical  lt  0              line completely missed         return false        else           var t1    -m b   Math sqrt underRadical    Math pow m 2    1     one of the intercept x s         var t2    -m b - Math sqrt underRadical    Math pow m 2    1     other intercept s x         var i1    x t1 c x  y m t1 b c y     intercept point 1         var i2    x t2 c x  y m t2 b c y     intercept point 2         return  i1  i2

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VB NET - Code  Function CheckLineSegmentCircleIntersection x1 As Double  y1 As Double  x2 As Double  y2 As Double  xc As Double  yc As Double  r As Double  As Boolean     Static xd As Double   0 0F     Static yd As Double   0 0F     Static t As Double   0 0F     Static d As Double   0 0F     Static dx 2 1 As Double   0 0F     Static dy 2 1 As Double   0 0F      dx 2 1   x2 - x1     dy 2 1   y2 - y1      t     yc - y1    dy 2 1    xc - x1    dx 2 1     dy 2 1   dy 2 1   dx 2 1   dx 2 1       If 0  lt   t And t  lt   1 Then         xd   x1   t   dx 2 1         yd   y1   t   dy 2 1          d   Math Sqrt  xd - xc     xd - xc     yd - yc     yd - yc           Return d  lt   r     Else         d   Math Sqrt  xc - x1     xc - x1     yc - y1     yc - y1           If d  lt   r Then             Return True         Else             d   Math Sqrt  xc - x2     xc - x2     yc - y2     yc - y2               If d  lt   r Then                 Return True             Else                 Return False             End If         End If     End If End Function

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I know it s been a while since this thread was open  From the answer given by chmike and improved by Aqib Mumtaz  They give a good answer but only works for a infinite line as said Aqib  So I add some comparisons to know if the line segment touch the circle  I write it in Python   def LineIntersectCircle c  r  p1  p2        p1 is the first line point      p2 is the second line point      c is the circle s center      r is the circle s radius      p3    p1 0 -c 0   p1 1 -c 1       p4    p2 0 -c 0   p2 1 -c 1        m    p4 1  - p3 1      p4 0  - p3 0       b   p3 1  - m   p3 0       underRadical   math pow r 2  math pow m 2    math pow r 2  - math pow b 2       if  underRadical  lt  0           print  NOT       else          t1    -2 m b 2 math sqrt underRadical      2   math pow m 2    2          t2    -2 m b-2 math sqrt underRadical      2   math pow m 2    2          i1    t1 c 0   m   t1   b   c 1           i2    t2 c 0   m   t2   b   c 1            if p1 0   gt  p2 0                                              Si el punto 1 es mayor al 2 en X             if  i1 0   lt  p1 0   and  i1 0   gt  p2 0                     Si el punto iX esta entre 2 y 1 en X                 if p1 1   gt  p2 1                                      Si el punto 1 es mayor al 2 en Y                     if  i1 1   lt  p1 1   and  i1 1   gt  p2 1             Si el punto iy esta entre 2 y 1                         print  Intersection                   if p1 1   lt  p2 1                                      Si el punto 2 es mayo al 2 en Y                     if  i1 1   gt  p1 1   and  i1 1   lt  p2 1             Si el punto iy esta entre 1 y 2                         print  Intersection            if p1 0   lt  p2 0                                              Si el punto 2 es mayor al 1 en X             if  i1 0   gt  p1 0   and  i1 0   lt  p2 0                     Si el punto iX esta entre 1 y 2 en X                 if p1 1   gt  p2 1                                      Si el punto 1 es mayor al 2 en Y                     if  i1 1   lt  p1 1   and  i1 1   gt  p2 1             Si el punto iy esta entre 2 y 1                         print  Intersection                   if p1 1   lt  p2 1                                      Si el punto 2 es mayo al 2 en Y                     if  i1 1   gt  p1 1   and  i1 1   lt  p2 1             Si el punto iy esta entre 1 y 2                         print  Intersection            if p1 0   gt  p2 0                                              Si el punto 1 es mayor al 2 en X             if  i2 0   lt  p1 0   and  i2 0   gt  p2 0                     Si el punto iX esta entre 2 y 1 en X                 if p1 1   gt  p2 1                                      Si el punto 1 es mayor al 2 en Y                     if  i2 1   lt  p1 1   and  i2 1   gt  p2 1             Si el punto iy esta entre 2 y 1                         print  Intersection                   if p1 1   lt  p2 1                                      Si el punto 2 es mayo al 2 en Y                     if  i2 1   gt  p1 1   and  i2 1   lt  p2 1             Si el punto iy esta entre 1 y 2                         print  Intersection            if p1 0   lt  p2 0                                              Si el punto 2 es mayor al 1 en X             if  i2 0   gt  p1 0   and  i2 0   lt  p2 0                     Si el punto iX esta entre 1 y 2 en X                 if p1 1   gt  p2 1                                      Si el punto 1 es mayor al 2 en Y                     if  i2 1   lt  p1 1   and  i2 1   gt  p2 1             Si el punto iy esta entre 2 y 1                         print  Intersection                   if p1 1   lt  p2 1                                      Si el punto 2 es mayo al 2 en Y                     if  i2 1   gt  p1 1   and  i2 1   lt  p2 1             Si el punto iy esta entre 1 y 2                         print  Intersection

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I wrote a small script to test intersection by projecting circle s center point on to line    vector distVector   centerPoint - projectedPoint  if distVector length    lt  circle radius        double distance   circle radius - distVector length        vector moveVector   distVector normalize     distance      circle move moveVector       http   jsfiddle net ercang ornh3594 1   If you need to check the collision with the segment  you also need to consider circle center s distance to start and end points   vector distVector   centerPoint - startPoint  if distVector length    lt  circle radius        double distance   circle radius - distVector length        vector moveVector   distVector normalize     distance      circle move moveVector       https   jsfiddle net ercang menp0991

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Solution in python  based on  Joe Skeen  def check line segment circle intersection line  point  radious           Checks whether a point intersects with a line defined by two points       A  point  is list with two values   2  3       A  line  is list with two points   point1  point2               line distance   distance line 0   line 1       distance start to point   distance line 0   point      distance end to point   distance line 1   point       if  distance start to point  lt   radious or distance end to point  lt   radious           return True        angle between line and point with law of cosines     numerator    math pow distance start to point  2                     math pow line distance  2                   - math pow distance end to point  2       denominator   2   distance start to point   line distance     ratio   numerator   denominator     ratio   ratio if ratio  lt   1 else 1    To account for float errors     ratio   ratio if ratio  gt   -1 else -1    To account for float errors     angle   math acos ratio         distance from the point to the line with sin projection     distance line to point   math sin angle    distance start to point      if distance line to point  lt   radious          point projection in line   math cos angle    distance start to point           Intersection occurs whent the point projection in the line is less           than the line distance and positive         return point projection in line  lt   line distance and point projection in line  gt   0     return False  def distance point1  point2       return math sqrt          math pow point1 1  - point2 1   2            math pow point1 0  - point2 0   2

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If the line s coordinates are A x  A y and B x  B y and the circles center is C x  C y then the lines formulae are    x   A x   t   B x    1 - t   y   A y   t   B y    1 - t   where 0 lt  t lt  1  and the circle is    C x - x  2    C y - y  2   R 2  if you substitute x and y formulae of the line into the circles formula you get a second order equation of t and its solutions are the intersection points  if there are any   If you get a t which is smaller than 0 or greater than 1 then its not a solution but it shows that the line is  pointing  to the direction of the circle

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Taking  E is the starting point of the ray  L is the end point of the ray  C is the center of sphere you re testing against r is the radius of that sphere  Compute  d   L - E   Direction vector of ray  from start to end   f   E - C   Vector from center sphere to ray start   Then the intersection is found by   Plugging  P   E   t   d This is a parametric equation  Px   Ex   tdx Py   Ey   tdy into  x - h 2    y - k 2   r2  h k    center of circle   Note  We ve simplified the problem to 2D here  the solution we get applies also in 3D  to get   Expand x2 - 2xh   h2   y2 - 2yk   k2 - r2   0 Plug x   ex   tdx y   ey   tdy   ex   tdx  2 - 2  ex   tdx  h   h2     ey   tdy  2 - 2  ey   tdy  k   k2 - r2   0 Explode ex2   2extdx   t2dx2 - 2exh - 2tdxh   h2   ey2   2eytdy   t2dy2 - 2eyk - 2tdyk   k2 - r2   0 Group t2  dx2   dy2     2t  exdx   eydy - dxh - dyk     ex2   ey2 - 2exh - 2eyk   h2   k2 - r2   0 Finally  t2  d    d     2t  e    d - d    c     e    e - 2  e    c     c    c - r2   0 Where d is the vector d and    is the dot product  And then  t2  d    d     2t  d      e - c         e - c        e - c   - r2   0 Letting f   e - c t2  d    d     2t  d    f     f    f - r2   0  So we get  t2    d    d    2t   f    d       f    f - r2     0 So solving the quadratic equation  float a   d Dot  d     float b   2 f Dot  d     float c   f Dot  f   - r r    float discriminant   b b-4 a c  if  discriminant  lt  0          no intersection   else        ray didn t totally miss sphere       so there is a solution to      the equation       discriminant   sqrt  discriminant          either solution may be on or off the ray so need to test both      t1 is always the smaller value  because BOTH discriminant and      a are nonnegative    float t1    -b - discriminant   2 a     float t2    -b   discriminant   2 a         3x HIT cases                -o- gt              -- -- gt                   -- - gt       Impale t1 hit t2 hit   Poke t1 hit t2 gt 1   ExitWound t1 lt 0  t2 hit          3x MISS cases             - gt   o                     o - gt                 - gt         FallShort  t1 gt 1 t2 gt 1   Past  t1 lt 0 t2 lt 0   CompletelyInside t1 lt 0  t2 gt 1       if  t1  gt   0  amp  amp  t1  lt   1              t1 is the intersection  and it s closer than t2         since t1 uses -b - discriminant         Impale  Poke     return true             here t1 didn t intersect so we are either started      inside the sphere or completely past it   if  t2  gt   0  amp  amp  t2  lt   1              ExitWound     return true               no intn  FallShort  Past  CompletelyInside   return false

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Okay  I won t give you code  but since you have tagged this algorithm  I don t think that will matter to you  First  you have to get a vector perpendicular to the line     You will have an unknown variable in y   ax   c   c will be unknown   To solve for that  Calculate it s value when the line passes through the center of the circle     That is  Plug in the location of the circle center to the line equation and solve for c  Then calculate the intersection point of the original line and its normal     This will give you the closest point on the line to the circle  Calculate the distance between this point and the circle center  using the magnitude of the vector   If this is less than the radius of the circle - voila  we have an intersection

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Circle is really a bad guy    So a good way is to avoid true circle  if you can  If you are doing collision check for games you can go with some simplifications and have just 3 dot products  and a few comparisons   I call this  fat point  or  thin circle   its kind of a ellipse with zero radius in a direction parallel to a segment  but full radius in a direction perpendicular to a segment  First  i would consider renaming and switching coordinate system to avoid excessive data   s0s1   B-A  s0qp   C-A  rSqr   r r    Second  index h in hvec2f means than vector must favor horisontal operations  like dot   det    Which means its components are to be placed in a separate xmm registers  to avoid shuffling hadd ing hsub ing  And here we go  with most performant version of simpliest collision detection for 2D game   bool fat point collides segment const hvec2f amp  s0qp  const hvec2f amp  s0s1  const float amp  rSqr        auto a   dot s0s1  s0s1         if  a    0      if you haven t zero-length segments omit this  as it would save you 1  mm comineq ss   instruction and 1 memory fetch               auto b   dot s0s1  s0qp           auto t   b   a     length of projection of s0qp onto s0s1           std  cout  lt  lt   t      lt  lt  t  lt  lt    n           if   t  gt   0   amp  amp   t  lt   1                             auto c   dot s0qp  s0qp               auto r2   c - a   t   t              return  r2  lt   rSqr      true if collides                        return false      I doubt you can optimize it any further  I am using it for neural-network driven car racing collision detection  to process millions of millions iteration steps

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Here is good solution in JavaScript  with all required mathematics and live illustration  https   bl ocks org milkbread 11000965  Though is on function in that solution needs modifications    x000D   x000D  function is on a  b  c    x000D      return Math abs distance a c    distance c b  - distance a b   lt 0 000001  x000D    x000D   x000D   x000D

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Maybe there is another way to solve this problem using rotation of coordinate system  Normally  if one segment is horizontal or vertical  which means parallel to x or y axis  it s quite easy to solve the intersection point since we already know one coordinate of the intersection  if any  The rest is obviously finding the other coordinate using circle s equation  Inspired by this idea  we could apply coordinates system rotation to make one axis s direction coincide with segment s direction  Let s take an example of circle x 2 y 2 1 and segment P1-P2 with P1 -1 5 0 5  and P2 -0 5 -0 5  in x-y system  And the following equations to remind you of the rotation principles  where theta is the angle anticlockwise  x -y  is the system after rotation    x    x   cos theta    y   sin theta  y    - x   sin theta    y   cos theta   and inversely  x   x    cos theta  - y    sin theta  y   x    sin theta    y    cos theta   Considering the segment P1-P2 direction  45   in terms of -x   we could take theta 45    Taking the second equations of rotation into circle s equation in x-y system   x 2 y 2 1 and after simple operations we get the  same  equation in x -y  system   x  2 y  2 1  Segment endpoints become in x -y  system using the first equations of rotation   gt  P1 -sqrt 2  2  sqrt 2    P2 -sqrt 2  2  0   Assuming the intersection as P  We have in x -y  Px   -sqrt 2  2  Using the new equation of circle  we get Py    sqrt 2  2  Converting P into original x-y system  we get finally P -1 0   To implement this numerically  we could firstly have a look at segment s direction   horizontal  vertical or not  If it belongs to the two first cases  it s simple like I said  If the last case  apply the algorithms above  To juge if there is intersection  we could compare the solution with the endpoints coordinates  to see whether there is one root between them   I believe this method could be also applied to other curves as long as we have its equation  The only weakness is that we should solve the equation in x -y  system for the other coordinate  which might be difficult

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If you find the distance between the center of the sphere  since it s 3D I assume you mean sphere and not circle  and the line  then check if that distance is less than the radius that will do the trick   The collision point is obviously the closest point between the line and the sphere  which will be calculated when you re calculating the distance between the sphere and the line   Distance between a point and a line  http   mathworld wolfram com Point-LineDistance3-Dimensional html

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I have created this function for iOS following the answer given by chmike     NSArray   intersectionPointsOfCircleWithCenter  CGPoint center withRadius  float radius toLinePoint1  CGPoint p1 andLinePoint2  CGPoint p2       NSMutableArray  intersectionPoints    NSMutableArray array        float Ax   p1 x      float Ay   p1 y      float Bx   p2 x      float By   p2 y      float Cx   center x      float Cy   center y      float R   radius           compute the euclidean distance between A and B     float LAB   sqrt  pow Bx-Ax  2  pow By-Ay  2             compute the direction vector D from A to B     float Dx    Bx-Ax  LAB      float Dy    By-Ay  LAB          Now the line equation is x   Dx t   Ax  y   Dy t   Ay with 0  lt   t  lt   1          compute the value t of the closest point to the circle center  Cx  Cy      float t   Dx  Cx-Ax    Dy  Cy-Ay           This is the projection of C on the line from A to B          compute the coordinates of the point E on line and closest to C     float Ex   t Dx Ax      float Ey   t Dy Ay          compute the euclidean distance from E to C     float LEC   sqrt  pow Ex-Cx  2   pow Ey-Cy  2             test if the line intersects the circle     if  LEC  lt  R                    compute distance from t to circle intersection point         float dt   sqrt  pow R  2  - pow LEC 2                 compute first intersection point         float Fx    t-dt  Dx   Ax          float Fy    t-dt  Dy   Ay              compute second intersection point         float Gx    t dt  Dx   Ax          float Gy    t dt  Dy   Ay            intersectionPoints addObject  NSValue valueWithCGPoint CGPointMake Fx  Fy              intersectionPoints addObject  NSValue valueWithCGPoint CGPointMake Gx  Gy                   else test if the line is tangent to circle     else if  LEC    R                tangent point to circle is E          intersectionPoints addObject  NSValue valueWithCGPoint CGPointMake Ex  Ey               else              line doesn t touch circle            return intersectionPoints

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In this post circle line collision will be checked by checking distance between circle center and point on line segment  Ipoint  that represent intersection point between normal N  Image 2  from circle center to line segment     https   i stack imgur com 3o6do png   On image 1 one circle and one line are shown  vector A point to line start point  vector B point to line end point  vector C point to circle center  Now we must find vector E  from line start point to circle center  and vector D  from line start point to line end point  this calculation is shown on image 1    https   i stack imgur com 7098a png    At image 2 we can see that vector E is projected on Vector D by  dot product  of vector E and unit vector D  result of dot product is scalar Xp that represent the distance between line start point and point of intersection  Ipoint  of vector N and vector D  Next vector X is found  by multiplying unit vector D and scalar Xp   Now we need to find vector Z  vector to Ipoint   its easy its simple vector addition of vector A  start point on line  and vector X  Next we need to deal with special cases we must check is Ipoint on line segment  if its not we must find out is it left of it or right of it  we will use vector closest to determine which point is closest to circle     https   i stack imgur com p9WIr png   When projection Xp is negative Ipoint is left of line segment  vector closest is equal to vector of line start point  when projection Xp is greater then magnitude of vector D then Ipoint is right of line segment  then closest vector is equal to vector of line end point in any other case closest vector is equal to vector Z   Now when we have closest vector   we need to find vector from circle center to Ipoint  dist vector   its simple we just need to subtract closest vector from center vector  Next just check if vector dist magnitude is less then circle radius if it is then they collide  if its not there is no collision    https   i stack imgur com QJ63q png   For end  we can return some values for resolving collision   easiest way is to return overlap of collision  subtract radius from vector dist magnitude  and return axis of collision  its vector D  Also intersection point is vector Z if needed

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Another one in c   partial Circle class   Tested and works like a charm   public class Circle   IEquatable lt Circle gt                                                                                     The center of a circle     private Point  center         The radius of a circle     private double  radius                                                                                     lt summary gt          Find all intersections  0  1  2  of the circle with a line defined by its 2 points          Using  http   math stackexchange com questions 228841 how-do-i-calculate-the-intersections-of-a-straight-line-and-a-circle         Note  p is the Center X and q is Center Y          lt  summary gt           lt param name  linePoint1  gt  lt  param gt           lt param name  linePoint2  gt  lt  param gt           lt returns gt  lt  returns gt      public List lt Point gt  GetIntersections Point linePoint1  Point linePoint2                List lt Point gt  intersections   new List lt Point gt              double dx   linePoint2 X - linePoint1 X           if  dx AboutEquals 0      Straight vertical line                       if  linePoint1 X AboutEquals Center X - Radius     linePoint1 X AboutEquals Center X   Radius                                 Point pt   new Point linePoint1 X  Center Y                   intersections Add pt                             else if  linePoint1 X  gt  Center X - Radius  amp  amp  linePoint1 X  lt  Center X   Radius                                double x   linePoint1 X - Center X                   Point pt   new Point linePoint1 X  Center Y   Math Sqrt Radius   Radius -  x   x                     intersections Add pt                    pt   new Point linePoint1 X  Center Y - Math Sqrt Radius   Radius -  x   x                     intersections Add pt                              return intersections                        Line function  y   mx   b          double dy   linePoint2 Y - linePoint1 Y          double m   dy   dx          double b   linePoint1 Y - m   linePoint1 X           double A   m   m   1          double B   2    m   b - m    center Y - Center X           double C   Center X   Center X   Center Y   Center Y - Radius   Radius - 2   b   Center Y   b   b           double discriminant   B   B - 4   A   C           if  discriminant  lt  0                        return intersections     there is no intersections                    if  discriminant AboutEquals 0      Tangeante  touch on 1 point only                        double x   -B    2   A               double y   m   x   b               intersections Add new Point x  y                      else    Secant  touch on 2 points                        double x    -B   Math Sqrt discriminant      2   A               double y   m   x   b              intersections Add new Point x  y                 x    -B - Math Sqrt discriminant      2   A               y   m   x   b              intersections Add new Point x  y                       return intersections                                                                                          Get the center      XmlElement  Center        public Point Center               get   return  center            set                        center   value                                                                                                    Get the radius      XmlElement      public double Radius               get   return  radius            set    radius   value                                                                                              XmlArrayItemAttribute  DoublePoint          public List lt Point gt  Coordinates                   get   return  coordinates                                                                                              Construct a circle without any specification     public Circle                  center X   0           center Y   0           radius   0                                                                                          Construct a circle without any specification     public Circle double radius                 center X   0           center Y   0           radius   radius                                                                                          Construct a circle with the specified circle     public Circle Circle circle                 center   circle  center           radius   circle  radius                                                                                          Construct a circle with the specified center and radius     public Circle Point center  double radius                 center   center           radius   radius                                                                                          Construct a circle based on one point     public Circle Point center                 center   center           radius   0                                                                                          Construct a circle based on two points     public Circle Point p1  Point p2                Circle2Points p1  p2           Required   using System   namespace Mathematic       public static class DoubleExtension                                                                                                Base on Hans Passant Answer on             http   stackoverflow com questions 2411392 double-epsilon-for-equality-greater-than-less-than-less-than-or-equal-to-gre               lt summary gt              Compare two double taking in account the double precision potential error              Take care  truncation errors accumulate on calculation  More you do  more you should increase the epsilon          public static bool AboutEquals this double value1  double value2                        if  double IsPositiveInfinity value1                   return double IsPositiveInfinity value2                if  double IsNegativeInfinity value1                   return double IsNegativeInfinity value2                if  double IsNaN value1                   return double IsNaN value2                double epsilon   Math Max Math Abs value1   Math Abs value2     1E-15              return Math Abs value1 - value2   lt   epsilon                                                                                                      Base on Hans Passant Answer on             http   stackoverflow com questions 2411392 double-epsilon-for-equality-greater-than-less-than-less-than-or-equal-to-gre               lt summary gt              Compare two double taking in account the double precision potential error              Take care  truncation errors accumulate on calculation  More you do  more you should increase the epsilon              You get really better performance when you can determine the contextual epsilon first               lt  summary gt               lt param name  value1  gt  lt  param gt               lt param name  value2  gt  lt  param gt               lt param name  precalculatedContextualEpsilon  gt  lt  param gt               lt returns gt  lt  returns gt          public static bool AboutEquals this double value1  double value2  double precalculatedContextualEpsilon                        if  double IsPositiveInfinity value1                   return double IsPositiveInfinity value2                if  double IsNegativeInfinity value1                   return double IsNegativeInfinity value2                if  double IsNaN value1                   return double IsNaN value2                return Math Abs value1 - value2   lt   precalculatedContextualEpsilon                                                                                                   public static double GetContextualEpsilon this double biggestPossibleContextualValue                        return biggestPossibleContextualValue   1E-15                                                                                                        lt summary gt              Mathlab equivalent              lt  summary gt               lt param name  dividend  gt  lt  param gt               lt param name  divisor  gt  lt  param gt               lt returns gt  lt  returns gt          public static double Mod this double dividend  double divisor                        return dividend - System Math Floor dividend   divisor    divisor

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You ll need some math here   Suppose A    Xa  Ya   B    Xb  Yb  and C    Xc  Yc   Any point on the line from A to B has coordinates  alpha Xa    1-alpha Xb  alphaYa    1-alpha  Yb    P  If the point P has distance R to C  it must be on the circle  What you want is to solve  distance P  C    R   that is   alpha Xa    1-alpha  Xb  2    alpha Ya    1-alpha  Yb  2   R 2 alpha 2 Xa 2   alpha 2 Xb 2 - 2 alpha Xb 2   Xb 2   alpha 2 Ya 2   alpha 2 Yb 2 - 2 alpha Yb 2   Yb 2 R 2  Xa 2   Xb 2   Ya 2   Yb 2  alpha 2 - 2  Xb 2   Yb 2  alpha    Xb 2   Yb 2 - R 2    0   if you apply the ABC-formula to this equation to solve it for alpha  and compute the coordinates of P using the solution s  for alpha  you get the intersection points  if any exist

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This solution I found seemed a little easier to follow then some of the other ones    Taking   p1 and p2 as the points for the line  and c as the center point for the circle and r for the radius   I would solve for the equation of the line in slope-intercept form  However  I didn t want to have to deal with difficult equations with c as a point  so I just shifted the coordinate system over so that the circle is at 0 0  p3   p1 - c p4   p2 - c   By the way  whenever I subtract points from each other I am subtracting the x s and then subtracting the y s  and putting them into a new point  just in case someone didn t know   Anyway  I now solve for the equation of the line with p3 and p4   m    p4 y - p3 y     p4 x - p3   the underscore is an attempt at subscript  y   mx   b y - mx   b  just put in a point for x and y  and insert the m we found    Ok  Now I need to set these equations equal  First I need to solve the circle s equation for x  x 2   y 2   r 2 y 2   r 2 - x 2 y   sqrt r 2 - x 2    Then I set them equal   mx   b   sqrt r 2 - x 2    And solve for the quadratic equation  0   ax 2   bx   c     mx   b  2   r 2 - x 2  mx  2   2mbx   b 2   r 2 - x 2 0   m 2   x 2   x 2   2mbx   b 2 - r 2 0    m 2   1    x 2   2mbx   b 2 - r 2   Now I have my a  b  and c   a   m 2   1 b   2mb c   b 2 - r 2   So I put this into the quadratic formula    -b    sqrt b 2 - 4ac     2a   And substitute in by values then simplify as much as possible    -2mb    sqrt b 2 - 4ac     2a  -2mb    sqrt  -2mb  2 - 4 m 2   1  b 2 - r 2      2 m 2   1   -2mb    sqrt 4m 2   b 2 - 4 m 2   b 2 - m 2   r 2   b 2 - r 2      2m 2   2  -2mb    sqrt 4    m 2   b 2 -  m 2   b 2 - m 2   r 2   b 2 - r 2      2m 2   2  -2mb    sqrt 4    m 2   b 2 - m 2   b 2   m 2   r 2 - b 2   r 2     2m 2   2  -2mb    sqrt 4    m 2   r 2 - b 2   r 2     2m 2   2  -2mb    sqrt 4    sqrt m 2   r 2 - b 2   r 2    2m 2   2  -2mb    2   sqrt m 2   r 2 - b 2   r 2    2m 2   2  -2mb    2   sqrt m 2   r 2   r 2 - b 2    2m 2   2  -2mb    2   sqrt r 2    m 2   1  - b 2    2m 2   2   This is almost as far as it will simplify  Finally  separate out to equations with the       -2mb   2   sqrt r 2    m 2   1  - b 2    2m 2   2 or       -2mb - 2   sqrt r 2    m 2   1  - b 2    2m 2   2    Then simply plug the result of both of those equations into the x in mx   b  For clarity  I wrote some JavaScript code to show how to use this   function interceptOnCircle p1 p2 c r         p1 is the first line point       p2 is the second line point       c is the circle s center       r is the circle s radius      var p3    x p1 x - c x  y p1 y - c y    shifted line points     var p4    x p2 x - c x  y p2 y - c y       var m    p4 y - p3 y     p4 x - p3 x     slope of the line     var b   p3 y - m   p3 x    y-intercept of line      var underRadical   Math pow  Math pow r 2   Math pow m 2  1   2 -Math pow b 2      the value under the square root sign       if  underRadical  lt  0         line completely missed         return false        else           var t1    -2 m b 2 Math sqrt underRadical    2   Math pow m 2    2     one of the intercept x s         var t2    -2 m b-2 Math sqrt underRadical    2   Math pow m 2    2     other intercept s x         var i1    x t1 y m t1 b    intercept point 1         var i2    x t2 y m t2 b    intercept point 2         return  i1 i2             I hope this helps   P S  If anyone finds any errors or has any suggestions  please comment  I am very new and welcome all help suggestions

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I just needed that  so I came up with this solution  The language is maxscript  but it should be easily translated to any other language  sideA  sideB and CircleRadius are scalars  the rest of the variables are points as  x y z   I m assuming z 0 to solve on the plane XY  fn projectPoint p1 p2 p3   --project  p1 perpendicular to the line p2-p3       local v  normalize  p3-p2      local p   p1-p2      p2   dot v p  v    fn findIntersectionLineCircle CircleCenter CircleRadius LineP1 LineP2        pp projectPoint CircleCenter LineP1 LineP2     sideA distance pp CircleCenter     --use pythagoras to solve the third side     sideB sqrt CircleRadius 2-sideA 2  -- this will return NaN if they don t intersect     IntersectV normalize  pp-CircleCenter      perpV  IntersectV y -IntersectV x IntersectV z      --project the point to both sides to find the solutions     solution1 pp  sideB perpV      solution2 pp- sideB perpV      return   solution1 solution2

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Here is a solution written in golang  The method is similar to some other answers posted here  but not quite the same  It is easy to implement  and has been tested  Here are the steps    Translate coordinates so that the circle is at the origin  Express the line segment as parametrized functions of t for both the x and y coordinates  If t is 0  the function s values are one end point of the segment  and if t is 1  the function s values are the other end point  Solve  if possible  the quadratic equation resulting from constraining values of t that produce x  y coordinates with distances from the origin equal to the circle s radius  Throw out solutions where t is  lt  0 or   1    lt   0 or    1 for an open segment   Those points are not contained in the segment  Translate back to original coordinates    The values for A  B  and C for the quadratic are derived here  where  n-et  and  m-dt  are the equations for the line s x and y coordinates  respectively  r is the radius of the circle    n-et  n-et     m-dt  m-dt    rr nn - 2etn   etet   mm - 2mdt   dtdt   rr  ee dd tt - 2 en   dm t   nn   mm - rr   0   Therefore A   ee dd  B   - 2 en   dm   and C   nn   mm - rr   Here is the golang code for the function   package geom  import        math        SegmentCircleIntersection return points of intersection between a circle and    a line segment  The Boolean intersects returns true if one or    more solutions exist  If only one solution exists      x1    x2 and y1    y2     s1x and s1y are coordinates for one end point of the segment  and    s2x and s2y are coordinates for the other end of the segment     cx and cy are the coordinates of the center of the circle and    r is the radius of the circle  func SegmentCircleIntersection s1x  s1y  s2x  s2y  cx  cy  r float64   x1  y1  x2  y2 float64  intersects bool            n-et  and  m-dt  are expressions for the x and y coordinates        of a parameterized line in coordinates whose origin is the        center of the circle         When t   0   n-et     s1x - cx and  m-dt     s1y - cy        When t   1   n-et     s2x - cx and  m-dt     s2y - cy      n    s2x - cx     m    s2y - cy      e    s2x - s1x     d    s2y - s1y         lineFunc checks if the  t parameter is in the segment and if so        calculates the line point in the unshifted coordinates  adds back        cx and cy      lineFunc    func t float64   x  y float64  inBounds bool            inBounds   t  gt   0  amp  amp  t  lt   1    Check bounds on closed segment            To check bounds for an open segment use t  gt  0  amp  amp  t  lt  1         if inBounds      Calc coords for point in segment             x   n - e t   cx             y   m - d t   cy                   return               Since we want the points on the line distance r from the origin          n-et  n-et     m-dt  m-dt    rr         Expanding and collecting terms yeilds the following quadratic equation      A  B  C    e e d d  -2  e n m d   n n m m-r r      D    B B - 4 A C    discriminant of quadratic     if D  lt  0           return    No solution           D   math Sqrt D       var p1In  p2In bool     x1  y1  p1In   lineFunc  -B   D     2   A      First root     if D    0 0           intersects   p1In         x2  y2   x1  y1         return    Only possible solution  quadratic has one root             x2  y2  p2In   lineFunc  -B - D     2   A      Second root      intersects   p1In    p2In     if p1In    false      Only x2  y2 may be valid solutions         x1  y1   x2  y2       else if p2In    false      Only x1  y1 are valid solutions         x2  y2   x1  y1           return     I tested it with this function  which confirms that solution points are within the line segment and on the circle  It makes a test segment and sweeps it around the given circle   package geom test  import        testing            put your package path here       func CheckEpsilon t  testing T  v  epsilon float64  message string        if v  gt  epsilon    v  lt  -epsilon           t Error message  v  epsilon          t FailNow            func TestSegmentCircleIntersection t  testing T        epsilon    1e-10         Something smallish     x1  y1    5 0  2 0       segment end point 1     x2  y2    50 0  30 0     segment end point 2     cx  cy    100 0  90 0    center of circle     r    80 0      segx  segy    x2-x1  y2-y1      testCntr  solutionCntr    0  0      for i    -100  i  lt  100  i             for j    -100  j  lt  100  j                 testCntr               s1x  s2x    x1 float64 i   x2 float64 i              s1y  s2y    y1 float64 j   y2 float64 j               sc1x  sc1y    s1x-cx  s1y-cy             seg1Inside    sc1x sc1x sc1y sc1y  lt  r r             sc2x  sc2y    s2x-cx  s2y-cy             seg2Inside    sc2x sc2x sc2y sc2y  lt  r r              p1x  p1y  p2x  p2y  intersects    SegmentCircleIntersection s1x  s1y  s2x  s2y  cx  cy  r               if intersects                   solutionCntr                     Check if points are on circle                 c1x  c1y    p1x-cx  p1y-cy                 deltaLen1     c1x c1x   c1y c1y  - r r                 CheckEpsilon t  deltaLen1  epsilon   p1 not on circle                    c2x  c2y    p2x-cx  p2y-cy                 deltaLen2     c2x c2x   c2y c2y  - r r                 CheckEpsilon t  deltaLen2  epsilon   p2 not on circle                       Check if points are on the line through the line segment                     cross product  of vector from a segment point to the point                    and the vector for the segment should be near zero                 vp1x  vp1y    p1x-s1x  p1y-s1y                 crossProd1    vp1x segy - vp1y segx                 CheckEpsilon t  crossProd1  epsilon   p1 not on line                     vp2x  vp2y    p2x-s1x  p2y-s1y                 crossProd2    vp2x segy - vp2y segx                 CheckEpsilon t  crossProd2  epsilon   p2 not on line                        Check if point is between points s1 and s2 on line                    This means the sign of the dot prod of the segment vector                    and point to segment end point vectors are opposite for                    either end                  wp1x  wp1y    p1x-s2x  p1y-s2y                 dp1v    vp1x segx   vp1y segy                 dp1w    wp1x segx   wp1y segy                 if  dp1v  lt  0  amp  amp  dp1w  lt  0      dp1v  gt  0  amp  amp  dp1w  gt  0                        t Error  point not contained in segment    dp1v  dp1w                      t FailNow                                      wp2x  wp2y    p2x-s2x  p2y-s2y                 dp2v    vp2x segx   vp2y segy                 dp2w    wp2x segx   wp2y segy                 if  dp2v  lt  0  amp  amp  dp2w  lt  0      dp2v  gt  0  amp  amp  dp2w  gt  0                        t Error  point not contained in segment    dp2v  dp2w                      t FailNow                                      if s1x    s2x  amp  amp  s2y    s1y     Only one solution                        Test that one end of the segment is withing the radius of the circle                        and one is not                     if seg1Inside  amp  amp  seg2Inside                           t Error  Only one solution but both line segment ends inside                           t FailNow                                             if  seg1Inside  amp  amp   seg2Inside                           t Error  Only one solution but both line segment ends outside                           t FailNow                                                          else      No intersection  check if both points outside or inside                 if  seg1Inside  amp  amp   seg2Inside       seg1Inside  amp  amp  seg2Inside                        t Error  No solution but only one point in radius of circle                       t FailNow                                                       t Log  Tested    testCntr    examples and found    solutionCntr    solutions        Here is the output of the test       RUN   TestSegmentCircleIntersection --- PASS  TestSegmentCircleIntersection  0 00s      geom test go 105  Tested  40000  examples and found  7343  solutions    Finally  the method is easily extendable to the case of a ray starting at one point  going through the other and extending to infinity  by only testing if t   0 or t  lt  1 but not both

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This Java Function returns a DVec2 Object  It takes a DVec2 for the center of the circle  the radius of the circle  and a Line   public static DVec2 CircLine DVec2 C  double r  Line line        DVec2 A   line p1      DVec2 B   line p2      DVec2 P      DVec2 AC   new DVec2  C        AC sub A       DVec2 AB   new DVec2  B        AB sub A       double ab2   AB dot AB       double acab   AC dot AB       double t   acab   ab2       if  t  lt  0 0           t   0 0      else if  t  gt  1 0           t   1 0         P   A   t   AB      P   new DVec2  AB        P mul  t        P add  A         DVec2 H   new DVec2  P        H sub  C        double h2   H dot H       double r2   r   r       if h2  gt  r2           return null      else         return P

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Just an addition to this thread    Below is a version of the code posted by pahlevan  but for C  XNA and tidied up a little            lt summary gt          Intersects a line and a circle           lt  summary gt           lt param name  location  gt the location of the circle lt  param gt           lt param name  radius  gt the radius of the circle lt  param gt           lt param name  lineFrom  gt the starting point of the line lt  param gt           lt param name  lineTo  gt the ending point of the line lt  param gt           lt returns gt true if the line and circle intersect each other lt  returns gt      public static bool IntersectLineCircle Vector2 location  float radius  Vector2 lineFrom  Vector2 lineTo                float ab2  acab  h2          Vector2 ac   location - lineFrom          Vector2 ab   lineTo - lineFrom          Vector2 Dot ref ab  ref ab  out ab2           Vector2 Dot ref ac  ref ab  out acab           float t   acab   ab2           if  t  lt  0              t   0          else if  t  gt  1              t   1           Vector2 h     ab   t    lineFrom  - location          Vector2 Dot ref h  ref h  out h2            return  h2  lt    radius   radius

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Here is my solution in TypeScript  following the idea that  Mizipzor suggested  using projection           Determines whether a line segment defined by a start and end point intersects with a sphere defined by a center point and a radius     param a the start point of the line segment     param b the end point of the line segment     param c the center point of the sphere     param r the radius of the sphere     export function lineSphereIntersects    a  IPoint    b  IPoint    c  IPoint    r  number    boolean        find the three sides of the triangle formed by the three points   const ab  number   distance a  b     const ac  number   distance a  c     const bc  number   distance b  c         check to see if either ends of the line segment are inside of the sphere   if  ac  lt  r    bc  lt  r        return true            find the angle between the line segment and the center of the sphere   const numerator  number   Math pow ac  2    Math pow ab  2  - Math pow bc  2     const denominator  number   2   ac   ab    const cab  number   Math acos numerator   denominator         find the distance from the center of the sphere and the line segment   const cd  number   Math sin cab    ac        if the radius is at least as long as the distance between the center and the line   if  r  gt   cd           find the distance between the line start and the point on the line closest to        the center of the sphere     const ad  number   Math cos cab    ac         intersection occurs when the point on the line closest to the sphere center is        no further away than the end of the line     return ad  lt   ab        return false     export function distance a  IPoint  b  IPoint   number     return Math sqrt      Math pow b z - a z  2    Math pow b y - a y  2    Math pow b x - a x  2          export interface IPoint     x  number    y  number    z  number

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Another method uses the triangle ABC area formula  The intersection test is simpler and more efficient than the projection method  but finding the coordinates of the intersection point requires more work  At least it will be delayed to the point it is required   The formula to compute the triangle area is   area   bh 2   where b is the base length and h is the height  We chose the segment AB to be the base so that h is the shortest distance from C  the circle center  to the line    Since the triangle area can also be computed by a vector dot product we can determine h      compute the triangle area times 2  area   area2 2  area2   abs   Bx-Ax   Cy-Ay  -  Cx-Ax  By-Ay        compute the AB segment length LAB   sqrt   Bx-Ax       By-Ay          compute the triangle height h   area2 LAB     if the line intersects the circle if  h  lt  R                         UPDATE 1    You could optimize the code by using the fast inverse square root computation described here to get a good approximation of 1 LAB   Computing the intersection point is not that difficult  Here it goes     compute the line AB direction vector components Dx    Bx-Ax  LAB Dy    By-Ay  LAB     compute the distance from A toward B of closest point to C t   Dx  Cx-Ax    Dy  Cy-Ay      t should be equal to sqrt   Cx-Ax       Cy-Ay    - h         compute the intersection point distance from t dt   sqrt  R   - h         compute first intersection point coordinate Ex   Ax    t-dt  Dx Ey   Ay    t-dt  Dy     compute second intersection point coordinate Fx   Ax    t dt  Dx Fy   Ay    t dt  Dy   If h   R then the line AB is tangent to the circle and the value dt   0 and E   F  The point coordinates are those of E and F   You should check that A is different of B and the segment length is not null if this may happen in your application

[algorithm] Circle line-segment collision detection algorithm?

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