几何学：从一个正方形的边缘到该正方形内可能的最大圆的边缘的距离

Question

数学从来都不是我的强项。：）给定一个任意大小的正方形画布，以像素为单位，我在该画布中绘制宽度为1的最大可能的圆。然后我想画一个矩形（不一定是正方形），在圆的中心与圆的角接触。所以我需要一个公式来计算左上角像素的X和Y坐标，给出圆的直径。或者，给定画布顶部的X个像素数和左侧的Y个像素数，我需要一个公式来找到该圆边缘上垂直和水平匹配点的X/Y坐标。谢了！：）

Answer 1

假定正弦规则适用，圆内正方形边的长度应该是Len=(Dia/sin(90))/2，那么左上角的X和Y坐标都是((dia-len)div2)。对于矩形，inc/dec等于X和Y。在Delphi中：

    procedure SquareInCircle(Dia: Integer; var P: TPoint; Padding: Integer = 0);
    //Given a circle of diameter Dia, find the X and Y of the
    //upper-left corner of a square with corners on the rim
    //of the circle.  Consider the diameter @ 45°, which forms
    //a right triangle from the square; the length of any side
    //of the square is the diameter / sin(90) / 2, and from that,
    //the X and Y of the upper-left are (dia - len) / 2.  Since
    //we're working in pixels, results are kept as integers.
    var
      L: Integer;
    begin
      L := Round((Dia / Sin(90)) / 2);
      P.X := ((Dia - L) div 2) + Padding;
      P.Y := P.X;
    end;

    procedure RectInCircle(Dia, Ofs: Integer; var P: TPoint; Padding: Integer = 0);
    //Ofs is the number of pixels to offset from square to rectangle.
    //If ofs is negative, the rectangle is taller than it is high;
    //if ofs is positive, the rectangle is wider than it is tall.
    //Ofs must be >= 0 and < (dia / 2).
    begin
      SquareInCircle(Dia, P, Padding);
      Dec(P.X, Ofs);
      Inc(P.Y, Ofs);
    end;