Formula, Circle's perimeter intersects with corners of rectangle

Question

Note: Everything about the circle is unknown. Only the width and height of the rectangle are known.

I devised a formula to determine the diameter of circle, so that the circle's perimeter intersects with the bottom two corners of a rectangle, whatever the proportion of the rectangle is (or at least if the width is greater than the height). The formula is as follows, whereby W refers to the width and H refers to the height of the rectangle:

X = W / H
Diameter = ((X/4) + (1/X)) * W

Does anybody know a more eloquent way to calculate this or is there a known algorithm to do this more efficiently?

Answer 1

Letting the center of the circle be (0, 0), the diameter be D, the width of the rectangle be W, and the height be H, the points (±W/2, D/2 − H) should lie on the circle, which by Pythagoras holds if and only if

(W/2)² + (D/2 − H)² = (D/2)².

Expanding the binomial and subtracting (D/2)² from both sides, we get

W²/4 − D H + H² = 0.

Solving for D by adding D H to both sides and dividing through by H, we get

D = W²/(4 H) + H.

In code this could be as follows.

double diameter(double width, double height) {
  return width * width / (4 * height) + height;
}