根据角度和直径定义CGPoint

Question

在drawRect 中定义了一个圆，下面的代码是用来定义圆的弧度。

CGFloat         width           = rect.size.width-rect.origin.x;
CGFloat         height          = rect.size.height-rect.origin.y;
CGFloat         xPos            = rect.origin.x;
CGFloat         yPos            = rect.origin.y;
CGFloat         arcStake        = (width * 2) * 0.25;
CGFloat         radius          = height/2;
CGPoint         centre          = CGPointMake(xPos+width/2, yPos+height/2);
CGFloat         angle           = acos(arcStake/(2*radius));
CGFloat         startAng        = radians(180) + angle;
CGFloat         endAng          = radians(360) - angle;

// Define 2 CGPoints of arc

CGMutablePathRef path = CGPathCreateMutable();
CGPathAddArc(path, NULL, centre.x, centre.y, radius, startAng, endAng, 0);
CGPathAddLineToPoint(path, NULL, xPos+width/2, yPos+height/2);
CGPathCloseSubpath(path);

定义两个CGPoints 的弧度，如下图的效果：

Answer 1

在半径为r的圆中，角度为a的点数有下面的坐标：
其中a是从圆的最右侧点开始测量。

x = r*cos(a) + center.x
y = r*sin(a) + center.y