这个函数检测Circle和Rectangle之间的碰撞(交集)。他的回答类似于e.James的方法，但这个方法检测矩形的所有角(不仅仅是右角)的碰撞。

注意:

aRect.origin.x和aRect.origin.y是矩形左下角的坐标!

aCircle。x和圆。y为圆心坐标!

static inline BOOL RectIntersectsCircle(CGRect aRect, Circle aCircle) {

    float testX = aCircle.x;
    float testY = aCircle.y;

    if (testX < aRect.origin.x)
        testX = aRect.origin.x;
    if (testX > (aRect.origin.x + aRect.size.width))
        testX = (aRect.origin.x + aRect.size.width);
    if (testY < aRect.origin.y)
        testY = aRect.origin.y;
    if (testY > (aRect.origin.y + aRect.size.height))
        testY = (aRect.origin.y + aRect.size.height);

    return ((aCircle.x - testX) * (aCircle.x - testX) + (aCircle.y - testY) * (aCircle.y - testY)) < aCircle.radius * aCircle.radius;
}

假设你有矩形的四条边，检查从这些边到圆心的距离，如果小于半径，那么这些形状是相交的。

if sqrt((rectangleRight.x - circleCenter.x)^2 +
        (rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleRight.x - circleCenter.x)^2 +
        (rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleLeft.x - circleCenter.x)^2 +
        (rectangleTop.y - circleCenter.y)^2) < radius
// then they intersect

if sqrt((rectangleLeft.x - circleCenter.x)^2 +
        (rectangleBottom.y - circleCenter.y)^2) < radius
// then they intersect

以下是我的做法:

bool intersects(CircleType circle, RectType rect)
{
    circleDistance.x = abs(circle.x - rect.x);
    circleDistance.y = abs(circle.y - rect.y);

    if (circleDistance.x > (rect.width/2 + circle.r)) { return false; }
    if (circleDistance.y > (rect.height/2 + circle.r)) { return false; }

    if (circleDistance.x <= (rect.width/2)) { return true; } 
    if (circleDistance.y <= (rect.height/2)) { return true; }

    cornerDistance_sq = (circleDistance.x - rect.width/2)^2 +
                         (circleDistance.y - rect.height/2)^2;

    return (cornerDistance_sq <= (circle.r^2));
}

下面是它的工作原理:

The first pair of lines calculate the absolute values of the x and y difference between the center of the circle and the center of the rectangle. This collapses the four quadrants down into one, so that the calculations do not have to be done four times. The image shows the area in which the center of the circle must now lie. Note that only the single quadrant is shown. The rectangle is the grey area, and the red border outlines the critical area which is exactly one radius away from the edges of the rectangle. The center of the circle has to be within this red border for the intersection to occur. The second pair of lines eliminate the easy cases where the circle is far enough away from the rectangle (in either direction) that no intersection is possible. This corresponds to the green area in the image. The third pair of lines handle the easy cases where the circle is close enough to the rectangle (in either direction) that an intersection is guaranteed. This corresponds to the orange and grey sections in the image. Note that this step must be done after step 2 for the logic to make sense. The remaining lines calculate the difficult case where the circle may intersect the corner of the rectangle. To solve, compute the distance from the center of the circle and the corner, and then verify that the distance is not more than the radius of the circle. This calculation returns false for all circles whose center is within the red shaded area and returns true for all circles whose center is within the white shaded area.

对于那些需要用SQL在地理坐标中计算圆/矩形碰撞的人， 这是我在oracle 11中实现的e.James建议算法。

在输入中，它需要圆坐标，圆半径km和矩形的两个顶点坐标:

CREATE OR REPLACE FUNCTION "DETECT_CIRC_RECT_COLLISION"
(
    circleCenterLat     IN NUMBER,      -- circle Center Latitude
    circleCenterLon     IN NUMBER,      -- circle Center Longitude
    circleRadius        IN NUMBER,      -- circle Radius in KM
    rectSWLat           IN NUMBER,      -- rectangle South West Latitude
    rectSWLon           IN NUMBER,      -- rectangle South West Longitude
    rectNELat           IN NUMBER,      -- rectangle North Est Latitude
    rectNELon           IN NUMBER       -- rectangle North Est Longitude
)
RETURN NUMBER
AS
    -- converts km to degrees (use 69 if miles)
    kmToDegreeConst     NUMBER := 111.045;

    -- Remaining rectangle vertices 
    rectNWLat   NUMBER;
    rectNWLon   NUMBER;
    rectSELat   NUMBER;
    rectSELon   NUMBER;

    rectHeight  NUMBER;
    rectWIdth   NUMBER;

    circleDistanceLat   NUMBER;
    circleDistanceLon   NUMBER;
    cornerDistanceSQ    NUMBER;

BEGIN
    -- Initialization of remaining rectangle vertices  
    rectNWLat := rectNELat;
    rectNWLon := rectSWLon;
    rectSELat := rectSWLat;
    rectSELon := rectNELon;

    -- Rectangle sides length calculation
    rectHeight := calc_distance(rectSWLat, rectSWLon, rectNWLat, rectNWLon);
    rectWidth := calc_distance(rectSWLat, rectSWLon, rectSELat, rectSELon);

    circleDistanceLat := abs( (circleCenterLat * kmToDegreeConst) - ((rectSWLat * kmToDegreeConst) + (rectHeight/2)) );
    circleDistanceLon := abs( (circleCenterLon * kmToDegreeConst) - ((rectSWLon * kmToDegreeConst) + (rectWidth/2)) );

    IF circleDistanceLon > ((rectWidth/2) + circleRadius) THEN
        RETURN -1;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLat > ((rectHeight/2) + circleRadius) THEN
        RETURN -1;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLon <= (rectWidth/2) THEN
        RETURN 0;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    IF circleDistanceLat <= (rectHeight/2) THEN
        RETURN 0;   --  -1 => NO Collision ; 0 => Collision Detected
    END IF;


    cornerDistanceSQ := POWER(circleDistanceLon - (rectWidth/2), 2) + POWER(circleDistanceLat - (rectHeight/2), 2);

    IF cornerDistanceSQ <=  POWER(circleRadius, 2) THEN
        RETURN 0;  --  -1 => NO Collision ; 0 => Collision Detected
    ELSE
        RETURN -1;  --  -1 => NO Collision ; 0 => Collision Detected
    END IF;

    RETURN -1;  --  -1 => NO Collision ; 0 => Collision Detected
END;    

我为处理形状创建了类 希望你喜欢

public class Geomethry {
  public static boolean intersectionCircleAndRectangle(int circleX, int circleY, int circleR, int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight){
    boolean result = false;

    float rectHalfWidth = rectangleWidth/2.0f;
    float rectHalfHeight = rectangleHeight/2.0f;

    float rectCenterX = rectangleX + rectHalfWidth;
    float rectCenterY = rectangleY + rectHalfHeight;

    float deltax = Math.abs(rectCenterX - circleX);
    float deltay = Math.abs(rectCenterY - circleY);

    float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;

    do{
        // check that distance between the centerse is more than the distance between the circumcircle of rectangle and circle
        if(lengthHypotenuseSqure > ((rectHalfWidth+circleR)*(rectHalfWidth+circleR) + (rectHalfHeight+circleR)*(rectHalfHeight+circleR))){
            //System.out.println("distance between the centerse is more than the distance between the circumcircle of rectangle and circle");
            break;
        }

        // check that distance between the centerse is less than the distance between the inscribed circle
        float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
        if(lengthHypotenuseSqure < ((rectMinHalfSide+circleR)*(rectMinHalfSide+circleR))){
            //System.out.println("distance between the centerse is less than the distance between the inscribed circle");
            result=true;
            break;
        }

        // check that the squares relate to angles
        if((deltax > (rectHalfWidth+circleR)*0.9) && (deltay > (rectHalfHeight+circleR)*0.9)){
            //System.out.println("squares relate to angles");
            result=true;
        }
    }while(false);

    return result;
}

public static boolean intersectionRectangleAndRectangle(int rectangleX, int rectangleY, int rectangleWidth, int rectangleHeight, int rectangleX2, int rectangleY2, int rectangleWidth2, int rectangleHeight2){
    boolean result = false;

    float rectHalfWidth = rectangleWidth/2.0f;
    float rectHalfHeight = rectangleHeight/2.0f;
    float rectHalfWidth2 = rectangleWidth2/2.0f;
    float rectHalfHeight2 = rectangleHeight2/2.0f;

    float deltax = Math.abs((rectangleX + rectHalfWidth) - (rectangleX2 + rectHalfWidth2));
    float deltay = Math.abs((rectangleY + rectHalfHeight) - (rectangleY2 + rectHalfHeight2));

    float lengthHypotenuseSqure = deltax*deltax + deltay*deltay;

    do{
        // check that distance between the centerse is more than the distance between the circumcircle
        if(lengthHypotenuseSqure > ((rectHalfWidth+rectHalfWidth2)*(rectHalfWidth+rectHalfWidth2) + (rectHalfHeight+rectHalfHeight2)*(rectHalfHeight+rectHalfHeight2))){
            //System.out.println("distance between the centerse is more than the distance between the circumcircle");
            break;
        }

        // check that distance between the centerse is less than the distance between the inscribed circle
        float rectMinHalfSide = Math.min(rectHalfWidth, rectHalfHeight);
        float rectMinHalfSide2 = Math.min(rectHalfWidth2, rectHalfHeight2);
        if(lengthHypotenuseSqure < ((rectMinHalfSide+rectMinHalfSide2)*(rectMinHalfSide+rectMinHalfSide2))){
            //System.out.println("distance between the centerse is less than the distance between the inscribed circle");
            result=true;
            break;
        }

        // check that the squares relate to angles
        if((deltax > (rectHalfWidth+rectHalfWidth2)*0.9) && (deltay > (rectHalfHeight+rectHalfHeight2)*0.9)){
            //System.out.println("squares relate to angles");
            result=true;
        }
    }while(false);

    return result;
  } 
}

预检查一个完全封装矩形的圆是否与矩形发生碰撞。 检查圆内的矩形角。 对于每条边，看看是否有一条线与圆相交。将中心点C投影到直线AB上，得到点d。如果CD的长度小于半径，则发生了碰撞。

    projectionScalar=dot(AC,AB)/(mag(AC)*mag(AB));
    if(projectionScalar>=0 && projectionScalar<=1) {
        D=A+AB*projectionScalar;
        CD=D-C;
        if(mag(CD)<circle.radius){
            // there was a collision
        }
    }