Equation for testing if a point is inside a circle

Question

If you have  a circle with center  center x  center y  and radius radius  how do you test if a given point with coordinates  x  y  is inside the circle

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You should check whether the distance from the center of the circle to the point is smaller than the radius using Python if  x-center x   2    y-center y   2  lt   radius  2        inside circle

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Find the distance between the center of the circle and the points given  If the distance between them is less than the radius then the point is inside the circle  if the distance between them is equal to the radius of the circle then the point is on the circumference of the circle  if the distance is greater than the radius then the point is outside the circle  int d   r 2 -   center x-x  2    center y-y  2    if d gt 0    print  quot inside quot    else if d  0    print  quot on the circumference quot    else   print  quot outside quot

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Calculate the Distance  D   Math Sqrt Math Pow center x - x  2    Math Pow center y - y  2   return D  lt   radius   that s in C    convert for use in python

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As said above -- use Euclidean distance   from math import hypot  def in radius c x  c y  r  x  y       return math hypot c x-x  c y-y   lt   r

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In general  x and y must satisfy  x - center x  2    y - center y  2  lt  radius 2   Please note that points that satisfy the above equation with  lt  replaced by    are considered the points on the circle  and the points that satisfy the above equation with  lt  replaced by  gt  are considered the outside the circle

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This is the same solution as mentioned by Jason Punyon  but it contains a pseudo-code example and some more details  I saw his answer after writing this  but I didn t want to remove mine   I think the most easily understandable way is to first calculate the distance between the circle s center and the point  I would use this formula   d   sqrt  circle x - x  2    circle y - y  2    Then  simply compare the result of that formula  the distance  d   with the radius  If the distance  d  is less than or equal to the radius  r   the point is inside the circle  on the edge of the circle if d and r are equal    Here is a pseudo-code example which can easily be converted to any programming language   function is in circle circle x  circle y  r  x  y        d   sqrt  circle x - x  2    circle y - y  2       return d  lt   r      Where circle x and circle y is the center coordinates of the circle  r is the radius of the circle  and x and y is the coordinates of the point

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As stated previously  to show if the point is in the circle we can use the following  if   x-center x  2    y - center y  2  lt  radius 2        in circle  lt -  True    else       in circle  lt -  False      To represent it graphically we can use   plot x  y  asp   1  xlim   c -1  1   ylim   c -1  1   col   ifelse  x-center x  2    y - center y  2  lt  radius 2  green   red    draw circle 0  0  1  nv   1000  border   NULL  col   NA  lty   1  lwd   1

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PHP  if     x -  center x     2     y -  center y     2   lt     radius   2        return true     Inside   else       return false     Outside

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As stated previously  to show if the point is in the circle we can use the following  if   x-center x  2    y - center y  2  lt  radius 2        in circle  lt -  True    else       in circle  lt -  False      To represent it graphically we can use   plot x  y  asp   1  xlim   c -1  1   ylim   c -1  1   col   ifelse  x-center x  2    y - center y  2  lt  radius 2  green   red    draw circle 0  0  1  nv   1000  border   NULL  col   NA  lty   1  lwd   1

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As said above -- use Euclidean distance   from math import hypot  def in radius c x  c y  r  x  y       return math hypot c x-x  c y-y   lt   r

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PHP  if     x -  center x     2     y -  center y     2   lt     radius   2        return true     Inside   else       return false     Outside

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In general  x and y must satisfy  x - center x  2    y - center y  2  lt  radius 2   Please note that points that satisfy the above equation with  lt  replaced by    are considered the points on the circle  and the points that satisfy the above equation with  lt  replaced by  gt  are considered the outside the circle

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I used the code below for beginners like me      public class incirkel    public static void main String   args        int x       int y       int middelx       int middely       int straal        Adjust the coordinates of x and y  x   -1  y   -2      Adjust the coordinates of the circle middelx   9   middely   9  straal    10           When x y is within the circle the message below will be printed     if     middelx - x     middelx - x                            middely - y     middely - y                          lt   straal   straal                             System out println  coordinaten x y vallen binnen cirkel          When x y is NOT within the circle the error message below will be printed       else           System err println  x y coordinaten vallen helaas buiten de cirkel

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You can use Pythagoras to measure the distance between your point and the centre and see if it s lower than the radius   def in circle center x  center y  radius  x  y       dist   math sqrt  center x - x     2    center y - y     2      return dist  lt   radius   EDIT  hat tip to Paul   In practice  squaring is often much cheaper than taking the square root and since we re only interested in an ordering  we can of course forego taking the square root   def in circle center x  center y  radius  x  y       square dist    center x - x     2    center y - y     2     return square dist  lt   radius    2   Also  Jason noted that  lt   should be replaced by  lt  and depending on usage this may actually make sense even though I believe that it s not true in the strict mathematical sense  I stand corrected

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You should check whether the distance from the center of the circle to the point is smaller than the radius using Python if  x-center x   2    y-center y   2  lt   radius  2        inside circle

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The equation below is a expression that tests if a point is within a given circle where xP  amp  yP are the coordinates of the point  xC  amp  yC are the coordinates of the center of the circle and R is the radius of that given circle      If the above expression is true then the point is within the circle   Below is a sample implementation in C        public static bool IsWithinCircle PointF pC  Point pP  Single fRadius           return Distance pC  pP   lt   fRadius             public static Single Distance PointF p1  PointF p2           Single dX   p1 X - p2 X          Single dY   p1 Y - p2 Y          Single multi   dX   dX   dY   dY          Single dist    Single Math Round  Single Math Sqrt multi   3            return  Single dist

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Calculate the Distance  D   Math Sqrt Math Pow center x - x  2    Math Pow center y - y  2   return D  lt   radius   that s in C    convert for use in python

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You should check whether the distance from the center of the circle to the point is smaller than the radius using Python if  x-center x   2    y-center y   2  lt   radius  2        inside circle

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Moving into the world of 3D if you want to check if a 3D point is in a Unit Sphere you end up doing something similar  All that is needed to work in 2D is to use 2D vector operations       public static bool Intersects Vector3 point  Vector3 center  float radius                Vector3 displacementToCenter   point - center           float radiusSqr   radius   radius           bool intersects   displacementToCenter magnitude  lt  radiusSqr           return intersects

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You should check whether the distance from the center of the circle to the point is smaller than the radius using Python if  x-center x   2    y-center y   2  lt   radius  2        inside circle

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You can use Pythagoras to measure the distance between your point and the centre and see if it s lower than the radius   def in circle center x  center y  radius  x  y       dist   math sqrt  center x - x     2    center y - y     2      return dist  lt   radius   EDIT  hat tip to Paul   In practice  squaring is often much cheaper than taking the square root and since we re only interested in an ordering  we can of course forego taking the square root   def in circle center x  center y  radius  x  y       square dist    center x - x     2    center y - y     2     return square dist  lt   radius    2   Also  Jason noted that  lt   should be replaced by  lt  and depending on usage this may actually make sense even though I believe that it s not true in the strict mathematical sense  I stand corrected

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Moving into the world of 3D if you want to check if a 3D point is in a Unit Sphere you end up doing something similar  All that is needed to work in 2D is to use 2D vector operations       public static bool Intersects Vector3 point  Vector3 center  float radius                Vector3 displacementToCenter   point - center           float radiusSqr   radius   radius           bool intersects   displacementToCenter magnitude  lt  radiusSqr           return intersects

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Mathematically  Pythagoras is probably a simple method as many have already mentioned    x-center x  2    y - center y  2  lt  radius 2   Computationally  there are quicker ways  Define   dx   abs x-center x  dy   abs y-center y  R   radius   If a point is more likely to be outside this circle then imagine a square drawn around it such that it s sides are tangents to this circle     if dx gt R then      return false  if dy gt R then      return false    Now imagine a square diamond drawn inside this circle such that it s vertices touch this circle   if dx   dy  lt   R then      return true    Now we have covered most of our space and only a small area of this circle remains in between our square and diamond to be tested  Here we revert to Pythagoras as above   if dx 2   dy 2  lt   R 2 then      return true else      return false    If a point is more likely to be inside this circle then reverse order of first 3 steps   if dx   dy  lt   R then      return true  if dx  gt  R then      return false  if dy  gt  R      then return false  if dx 2   dy 2  lt   R 2 then      return true else     return false      Alternate methods imagine a square inside this circle instead of a diamond but this requires slightly more tests and calculations with no computational advantage  inner square and diamonds have identical areas    k   R sqrt 2  if dx  lt   k and dy  lt   k then      return true    Update   For those interested in performance I implemented this method in c  and compiled with -O3   I obtained execution times by time   a out  I implemented this method  a normal method and a dummy method to determine timing overhead   Normal  21 3s This  19 1s Overhead  16 5s  So  it seems this method is more efficient in this implementation      compile gcc -O3  lt filename gt  c    run  time   a out   include  lt stdio h gt   include  lt stdlib h gt    define TRUE   0  0   define FALSE  0  1    define ABS x     x  lt 0   0- x    x    int xo  yo  R   int inline inCircle  int x  int y        19 1  19 1  19 1   int dx   ABS x-xo     if      dx  gt   R   return FALSE    int dy   ABS y-yo     if      dy  gt   R   return FALSE    if   dx dy  lt   R   return TRUE    return   dx dx   dy dy  lt   R R       int inline inCircleN  int x  int y        21 3  21 1  21 5   int dx   ABS x-xo     int dy   ABS y-yo     return   dx dx   dy dy  lt   R R       int inline dummy  int x  int y        16 6  16 5  16 4   int dx   ABS x-xo     int dy   ABS y-yo     return FALSE      define N 1000000000  int main      int x  y    xo   rand   1000  yo   rand   1000  R   1    int n   0    int c    for  c 0  c lt N  c         x   rand   1000  y   rand   1000        if   inCircle x y          if   inCircleN x y           if   dummy x y           n                printf    d of  d inside circle n   n  N

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Here is the simple java code for solving this problem    and the math behind it   https   math stackexchange com questions 198764 how-to-know-if-a-point-is-inside-a-circle  boolean insideCircle int   point  int   center  int radius        return  float Math sqrt  int Math pow point 0 -center 0  2   int Math pow point 1 -center 1  2    lt   radius

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My answer in C  as a complete cut  amp  paste  not optimized  solution   public static bool PointIsWithinCircle double circleRadius  double circleCenterPointX  double circleCenterPointY  double pointToCheckX  double pointToCheckY        return  Math Pow pointToCheckX - circleCenterPointX  2    Math Pow pointToCheckY - circleCenterPointY  2    lt   Math Pow circleRadius  2        Usage   if   PointIsWithinCircle 3  3  3   5   5

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As said above -- use Euclidean distance   from math import hypot  def in radius c x  c y  r  x  y       return math hypot c x-x  c y-y   lt   r

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Here is the simple java code for solving this problem    and the math behind it   https   math stackexchange com questions 198764 how-to-know-if-a-point-is-inside-a-circle  boolean insideCircle int   point  int   center  int radius        return  float Math sqrt  int Math pow point 0 -center 0  2   int Math pow point 1 -center 1  2    lt   radius

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Calculate the Distance  D   Math Sqrt Math Pow center x - x  2    Math Pow center y - y  2   return D  lt   radius   that s in C    convert for use in python

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Find the distance between the center of the circle and the points given  If the distance between them is less than the radius then the point is inside the circle  if the distance between them is equal to the radius of the circle then the point is on the circumference of the circle  if the distance is greater than the radius then the point is outside the circle  int d   r 2 -   center x-x  2    center y-y  2    if d gt 0    print  quot inside quot    else if d  0    print  quot on the circumference quot    else   print  quot outside quot

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The equation below is a expression that tests if a point is within a given circle where xP  amp  yP are the coordinates of the point  xC  amp  yC are the coordinates of the center of the circle and R is the radius of that given circle      If the above expression is true then the point is within the circle   Below is a sample implementation in C        public static bool IsWithinCircle PointF pC  Point pP  Single fRadius           return Distance pC  pP   lt   fRadius             public static Single Distance PointF p1  PointF p2           Single dX   p1 X - p2 X          Single dY   p1 Y - p2 Y          Single multi   dX   dX   dY   dY          Single dist    Single Math Round  Single Math Sqrt multi   3            return  Single dist

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My answer in C  as a complete cut  amp  paste  not optimized  solution   public static bool PointIsWithinCircle double circleRadius  double circleCenterPointX  double circleCenterPointY  double pointToCheckX  double pointToCheckY        return  Math Pow pointToCheckX - circleCenterPointX  2    Math Pow pointToCheckY - circleCenterPointY  2    lt   Math Pow circleRadius  2        Usage   if   PointIsWithinCircle 3  3  3   5   5

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Mathematically  Pythagoras is probably a simple method as many have already mentioned    x-center x  2    y - center y  2  lt  radius 2   Computationally  there are quicker ways  Define   dx   abs x-center x  dy   abs y-center y  R   radius   If a point is more likely to be outside this circle then imagine a square drawn around it such that it s sides are tangents to this circle     if dx gt R then      return false  if dy gt R then      return false    Now imagine a square diamond drawn inside this circle such that it s vertices touch this circle   if dx   dy  lt   R then      return true    Now we have covered most of our space and only a small area of this circle remains in between our square and diamond to be tested  Here we revert to Pythagoras as above   if dx 2   dy 2  lt   R 2 then      return true else      return false    If a point is more likely to be inside this circle then reverse order of first 3 steps   if dx   dy  lt   R then      return true  if dx  gt  R then      return false  if dy  gt  R      then return false  if dx 2   dy 2  lt   R 2 then      return true else     return false      Alternate methods imagine a square inside this circle instead of a diamond but this requires slightly more tests and calculations with no computational advantage  inner square and diamonds have identical areas    k   R sqrt 2  if dx  lt   k and dy  lt   k then      return true    Update   For those interested in performance I implemented this method in c  and compiled with -O3   I obtained execution times by time   a out  I implemented this method  a normal method and a dummy method to determine timing overhead   Normal  21 3s This  19 1s Overhead  16 5s  So  it seems this method is more efficient in this implementation      compile gcc -O3  lt filename gt  c    run  time   a out   include  lt stdio h gt   include  lt stdlib h gt    define TRUE   0  0   define FALSE  0  1    define ABS x     x  lt 0   0- x    x    int xo  yo  R   int inline inCircle  int x  int y        19 1  19 1  19 1   int dx   ABS x-xo     if      dx  gt   R   return FALSE    int dy   ABS y-yo     if      dy  gt   R   return FALSE    if   dx dy  lt   R   return TRUE    return   dx dx   dy dy  lt   R R       int inline inCircleN  int x  int y        21 3  21 1  21 5   int dx   ABS x-xo     int dy   ABS y-yo     return   dx dx   dy dy  lt   R R       int inline dummy  int x  int y        16 6  16 5  16 4   int dx   ABS x-xo     int dy   ABS y-yo     return FALSE      define N 1000000000  int main      int x  y    xo   rand   1000  yo   rand   1000  R   1    int n   0    int c    for  c 0  c lt N  c         x   rand   1000  y   rand   1000        if   inCircle x y          if   inCircleN x y           if   dummy x y           n                printf    d of  d inside circle n   n  N

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boolean isInRectangle double centerX  double centerY  double radius       double x  double y            return x  gt   centerX - radius  amp  amp  x  lt   centerX   radius  amp  amp               y  gt   centerY - radius  amp  amp  y  lt   centerY   radius           test if coordinate  x  y  is within a radius from coordinate  center x  center y  public boolean isPointInCircle double centerX  double centerY       double radius  double x  double y        if isInRectangle centerX  centerY  radius  x  y                 double dx   centerX - x          double dy   centerY - y          dx    dx          dy    dy          double distanceSquared   dx   dy          double radiusSquared   radius   radius          return distanceSquared  lt   radiusSquared            return false      This is more efficient  and readable  It avoids the costly square root operation   I also added a check to determine if the point is within the bounding rectangle of the circle   The rectangle check is unnecessary except with many points or many circles   If most points are inside circles  the bounding rectangle check will actually make things slower   As always  be sure to consider your use case

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In general  x and y must satisfy  x - center x  2    y - center y  2  lt  radius 2   Please note that points that satisfy the above equation with  lt  replaced by    are considered the points on the circle  and the points that satisfy the above equation with  lt  replaced by  gt  are considered the outside the circle

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In general  x and y must satisfy  x - center x  2    y - center y  2  lt  radius 2   Please note that points that satisfy the above equation with  lt  replaced by    are considered the points on the circle  and the points that satisfy the above equation with  lt  replaced by  gt  are considered the outside the circle

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This is the same solution as mentioned by Jason Punyon  but it contains a pseudo-code example and some more details  I saw his answer after writing this  but I didn t want to remove mine   I think the most easily understandable way is to first calculate the distance between the circle s center and the point  I would use this formula   d   sqrt  circle x - x  2    circle y - y  2    Then  simply compare the result of that formula  the distance  d   with the radius  If the distance  d  is less than or equal to the radius  r   the point is inside the circle  on the edge of the circle if d and r are equal    Here is a pseudo-code example which can easily be converted to any programming language   function is in circle circle x  circle y  r  x  y        d   sqrt  circle x - x  2    circle y - y  2       return d  lt   r      Where circle x and circle y is the center coordinates of the circle  r is the radius of the circle  and x and y is the coordinates of the point

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Calculate the Distance  D   Math Sqrt Math Pow center x - x  2    Math Pow center y - y  2   return D  lt   radius   that s in C    convert for use in python

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You can use Pythagoras to measure the distance between your point and the centre and see if it s lower than the radius   def in circle center x  center y  radius  x  y       dist   math sqrt  center x - x     2    center y - y     2      return dist  lt   radius   EDIT  hat tip to Paul   In practice  squaring is often much cheaper than taking the square root and since we re only interested in an ordering  we can of course forego taking the square root   def in circle center x  center y  radius  x  y       square dist    center x - x     2    center y - y     2     return square dist  lt   radius    2   Also  Jason noted that  lt   should be replaced by  lt  and depending on usage this may actually make sense even though I believe that it s not true in the strict mathematical sense  I stand corrected

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You can use Pythagoras to measure the distance between your point and the centre and see if it s lower than the radius   def in circle center x  center y  radius  x  y       dist   math sqrt  center x - x     2    center y - y     2      return dist  lt   radius   EDIT  hat tip to Paul   In practice  squaring is often much cheaper than taking the square root and since we re only interested in an ordering  we can of course forego taking the square root   def in circle center x  center y  radius  x  y       square dist    center x - x     2    center y - y     2     return square dist  lt   radius    2   Also  Jason noted that  lt   should be replaced by  lt  and depending on usage this may actually make sense even though I believe that it s not true in the strict mathematical sense  I stand corrected

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I used the code below for beginners like me      public class incirkel    public static void main String   args        int x       int y       int middelx       int middely       int straal        Adjust the coordinates of x and y  x   -1  y   -2      Adjust the coordinates of the circle middelx   9   middely   9  straal    10           When x y is within the circle the message below will be printed     if     middelx - x     middelx - x                            middely - y     middely - y                          lt   straal   straal                             System out println  coordinaten x y vallen binnen cirkel          When x y is NOT within the circle the error message below will be printed       else           System err println  x y coordinaten vallen helaas buiten de cirkel

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boolean isInRectangle double centerX  double centerY  double radius       double x  double y            return x  gt   centerX - radius  amp  amp  x  lt   centerX   radius  amp  amp               y  gt   centerY - radius  amp  amp  y  lt   centerY   radius           test if coordinate  x  y  is within a radius from coordinate  center x  center y  public boolean isPointInCircle double centerX  double centerY       double radius  double x  double y        if isInRectangle centerX  centerY  radius  x  y                 double dx   centerX - x          double dy   centerY - y          dx    dx          dy    dy          double distanceSquared   dx   dy          double radiusSquared   radius   radius          return distanceSquared  lt   radiusSquared            return false      This is more efficient  and readable  It avoids the costly square root operation   I also added a check to determine if the point is within the bounding rectangle of the circle   The rectangle check is unnecessary except with many points or many circles   If most points are inside circles  the bounding rectangle check will actually make things slower   As always  be sure to consider your use case

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As said above -- use Euclidean distance   from math import hypot  def in radius c x  c y  r  x  y       return math hypot c x-x  c y-y   lt   r

[algorithm] Equation for testing if a point is inside a circle

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