Suppose you had a square intersecting a circle like in Circular Logic, where one vertex is inside the circle and three are outside.

If the vertex inside the circle was not located at the center, but anywhere inside the circle, such that segments ending at that vertex were of length a and b and the circle had radius r, what would be the area of their intersection be?

Note: A geometric solution is expected, without using calculus.

(In reply to can this be simplified? -- spoiler by Charlie)

What if the circle intersects the other two sides of the square in two points per side?

Posted by Bractals on 2006-08-09 13:30:22