Circle - Axis Alined Rectangle Intersection - c++

Question

I want to create a function which takes a rectangle and a circle returns a boolean as to whether or not they intersect. What's the most efficient and simple way of doing this? The function looks something like this:
bool intersect(rectX, rectY, rectWidth, rectHeight, circleX, circleY, radius)
{
bool intersect;
//code I need
return intersect;
}
Please help me find the code I need. Thanks!

Answer 1

Your answer is at last, but I also want to inform you/future viewer, the following.
When you are denoting a rectangle by one point AND height-width that is not specific rectangle on axis-system, that is general form of a rectangle can be draw anywhere.
So, In your case: Rectangle can be draw in any direction(to make it away from circle, eg: UP,LEFT,RIGHT,BOTTOM etc. side to rectX,rectY). That's why here two possibilities generates AND your possibility will depends on what you need:
[A] 100% assurance of intersection, no matter how you draw rectangle
[b] At least one way to draw the rectangle so that it will intersect with circle
Case A:
bool assuredIntersect(rectX, rectY, rectWidth, rectHeight, circleX, circleY, radius){
bool intersect;
float distance=((rectX-circleX)^2+(rectY-circleY)^2)^0.5;
intersect=(radious>=distance);
return intersect;
}
Case B:
bool canIntersect(rectX, rectY, rectWidth, rectHeight, circleX, circleY, radius){
bool intersect;
float distance=((rectX-circleX)^2+(rectY-circleY)^2)^0.5;
float diagonal=((rectX+rectHeight)^2+(rectY-rectWidth)^2)^0.5;
intersect=((radious+diagonal)<distance);
return intersect;
}
But, if you are denoting rectangle by two-points-with-one-side (rectX1,rectY1 AND rectX2,rectY2 AND height or width). Then you can denote a rectangle specifically.
Note: If direction of rectangle is fixed(like shorter-side-from-point(rectX,rectY) is perpendicular-or-at-an-angle to x-axis) then also rectangle become specific as we can calculate (rectX2,rectY2). Eg: if angle is 90 then second point will be (rectX+rectHeight, rectY+rectWidth).
If we have function parameter like this:
#include <math.h>
#define PI 3.14159265
bool intersect(rX1, rY1, rX2, rY2, rAngle, cX, cY, cR){
//can be `intersect(rX1,rY1,rH,rW,rAngle, cX,cY, cR)`, and calculate rX2,rY2
//can be `intersect(rX1,rY1,rX2,rY2,rH, cX,cY, cR)`, and calculate rAngle
bool intersect;
//assume (rX1,rY1) as origin AND rectangle`s-side attached to this point is on both axis,
//THEN we need to recalculate coordinates according to this assumption
rAngle=rAngle*PI/180; //angle in radian
//NOTE: if in place of rAngle, rHeight or rWidth is given then you can calculate rAngle by trigonometry.
//moving rX1,xY1 to (0,0)
cX=cX-rX1; cY=cY-rY1;
rX2-=rX1; rY2-=rY1; rX1=rY1=0;
//rotating axis, rectangle, circle...
float cosA=cos(rAngle), sinA=sin(rAngle);
float tempX= cosA*rX2 + sinA*rY2;
float tempY= sinA*rX2 + cosA*rY2;
rX2=tempX; rY2=tempY;
tempX=cosA*cX + sinA*cY;
tempY=sinA*cX + cosA*cY;
cX=tempX; cY=tempY;
rX1-=cR;rY1-=cR; //enlarging(creating) virtual rectangle around original; After this...
rX2+=cR;rY2+=cR; //...if circle centre is inside this rectangle it will intersect with original rectangle
intersect=(cX<=rX2 && cX>=rX1 && cY<=rY2 && cY>=rY1);
return intersect;
}
point(x,y) rotation
# So, if you don't know angle then you can consider first two case.
[ANSWER] If rectangle is axis aligned then function will be:
bool intersect(rectX, rectY, rectWidth, rectHeight, circleX, circleY, radius){
bool intersect;
//calculating rX2,xY2
rX2=rectX + rectWidth; rY2=rectY + rectHeight;
rectX-=radius;rectY-=radius; //enlarging(creating) virtual rectangle around original; After this...
rX2+=radius;rY2+=radius; //...if circle centre is inside this rectangle it will intersect with original rectangle ...
intersect=(circleX<=rX2 && circleX>=rectX && circleY<=rY2 && circleY>=rectY);
return intersect;
}