圆与矩形相交只有两种情况:

圆的中心在矩形的内部，或者 矩形的一条边在圆上有一个点。

注意，这并不要求矩形与轴平行。

(一种方法是:如果没有一条边在圆中有点(如果所有的边都完全“在”圆外)，那么圆仍然可以与多边形相交的唯一方法是它完全位于多边形内部。)

有了这样的见解，就可以像下面这样工作，其中圆的中心是P，半径是R，矩形的顶点是A, B, C, D(不完整的代码):

def intersect(Circle(P, R), Rectangle(A, B, C, D)):
    S = Circle(P, R)
    return (pointInRectangle(P, Rectangle(A, B, C, D)) or
            intersectCircle(S, (A, B)) or
            intersectCircle(S, (B, C)) or
            intersectCircle(S, (C, D)) or
            intersectCircle(S, (D, A)))

如果你在写任何几何，你的库中可能已经有了上面的函数。否则，pointInRectangle()可以用几种方式实现;任何一般的多边形点方法都可以工作，但对于矩形，你可以检查这是否有效:

0 ≤ AP·AB ≤ AB·AB and 0 ≤ AP·AD ≤ AD·AD

intersectCircle()也很容易实现:一种方法是检查从P到直线的垂线的脚是否足够近并且在端点之间，否则检查端点。

最酷的是，同样的想法不仅适用于矩形，而且适用于一个圆与任何简单多边形的交点——甚至不必是凸多边形!

圆与矩形相交只有两种情况:

圆的中心在矩形的内部，或者 矩形的一条边在圆上有一个点。

注意，这并不要求矩形与轴平行。

(一种方法是:如果没有一条边在圆中有点(如果所有的边都完全“在”圆外)，那么圆仍然可以与多边形相交的唯一方法是它完全位于多边形内部。)

有了这样的见解，就可以像下面这样工作，其中圆的中心是P，半径是R，矩形的顶点是A, B, C, D(不完整的代码):

def intersect(Circle(P, R), Rectangle(A, B, C, D)):
    S = Circle(P, R)
    return (pointInRectangle(P, Rectangle(A, B, C, D)) or
            intersectCircle(S, (A, B)) or
            intersectCircle(S, (B, C)) or
            intersectCircle(S, (C, D)) or
            intersectCircle(S, (D, A)))

如果你在写任何几何，你的库中可能已经有了上面的函数。否则，pointInRectangle()可以用几种方式实现;任何一般的多边形点方法都可以工作，但对于矩形，你可以检查这是否有效:

0 ≤ AP·AB ≤ AB·AB and 0 ≤ AP·AD ≤ AD·AD

intersectCircle()也很容易实现:一种方法是检查从P到直线的垂线的脚是否足够近并且在端点之间，否则检查端点。

最酷的是，同样的想法不仅适用于矩形，而且适用于一个圆与任何简单多边形的交点——甚至不必是凸多边形!

如果你对一个更图形化的解决方案感兴趣，甚至在(平面上)旋转的矩形..

演示:https://jsfiddle.net/exodus4d/94mxLvqh/2691/

这个想法是:

将场景转换为原点[0,0] 如果矩形不在平面上，则旋转中心应在 (0,0) 将场景旋转回平面 计算交点

const hasIntersection = ({x: cx, y: cy, r: cr}, {x, y, width, height}) => { const distX = Math.abs(cx - x - width / 2); const distY = Math.abs(cy - y - height / 2); if (distX > (width / 2 + cr)) { return false; } if (distY > (height / 2 + cr)) { return false; } if (distX <= (width / 2)) { return true; } if (distY <= (height / 2)) { return true; } const Δx = distX - width / 2; const Δy = distY - height / 2; return Δx * Δx + Δy * Δy <= cr * cr; }; const rect = new DOMRect(50, 20, 100, 50); const circ1 = new DOMPoint(160, 80); circ1.r = 20; const circ2 = new DOMPoint(80, 95); circ2.r = 20; const canvas = document.getElementById('canvas'); const ctx = canvas.getContext('2d'); ctx.strokeRect(rect.x, rect.y, rect.width, rect.height); ctx.beginPath(); ctx.strokeStyle = hasIntersection(circ1, rect) ? 'red' : 'green'; ctx.arc(circ1.x, circ1.y, circ1.r, 0, 2 * Math.PI); ctx.stroke(); ctx.beginPath(); ctx.strokeStyle = hasIntersection(circ2, rect) ? 'red' : 'green'; ctx.arc(circ2.x, circ2.y, circ2.r, 0, 2 * Math.PI); ctx.stroke(); <canvas id="canvas"></canvas>

提示:而不是旋转矩形(4点)。你可以向相反的方向旋转圆(1点)。

有一种非常简单的方法来做到这一点，你必须在x和y上夹住一个点，但在正方形内部，当圆心在一个垂直轴上的两个正方形边界点之间时，你需要将这些坐标夹到平行轴上，只是要确保夹住的坐标不超过正方形的限制。 然后只需得到圆心与夹紧坐标之间的距离，并检查距离是否小于圆的半径。

以下是我是如何做到的(前4个点是方坐标，其余是圆点):

bool DoesCircleImpactBox(float x, float y, float x1, float y1, float xc, float yc, float radius){
    float ClampedX=0;
    float ClampedY=0;
    
    if(xc>=x and xc<=x1){
    ClampedX=xc;
    }
    
    if(yc>=y and yc<=y1){
    ClampedY=yc;
    }
    
    radius = radius+1;
    
    if(xc<x) ClampedX=x;
    if(xc>x1) ClampedX=x1-1;
    if(yc<y) ClampedY=y;
    if(yc>y1) ClampedY=y1-1;
    
    float XDif=ClampedX-xc;
    XDif=XDif*XDif;
    float YDif=ClampedY-yc;
    YDif=YDif*YDif;
    
    if(XDif+YDif<=radius*radius) return true;
    
    return false;
}

为我工作(只工作时，矩形的角度是180)

function intersects(circle, rect) {
  let left = rect.x + rect.width > circle.x - circle.radius;
  let right = rect.x < circle.x + circle.radius;
  let top = rect.y < circle.y + circle.radius;
  let bottom = rect.y + rect.height > circle.y - circle.radius;
  return left && right && bottom && top;
}

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;
  } 
}