You draw a square. Then, you draw the largest possible circle, tangent to all four sides. If the lower left corner of the square is at (0,0), and the sides of the square are parallel to the coordinate axis, the point (2,1) is on the circle, within the square.

What is the radius of the circle?

The equation of a circle is (X - h)² + (Y - k)² = r², where h and k are the x-coordinate and y-coordinate of the center of the circle of radius r.
As the problem gives the circle is tangent to all four sides of the square which has its bottom left corner at coordinate (0,0), it can be established that h and k are equal to the radius r and, thus, the equation can be written as (X - r)² + (Y - r)² = r².
Given that (2,1) is a point on the circle we can substitute X and Y with the values 2 and 1. Simplifying, the resulting equation can be written as r² - 6r + 5 = 0. Applying the quadratic formula, two real results are found: 1 and 5.  As a solution of radius 1 would put the point (2,1) on the circle where it is tangent to the right side of the square, and not within the square, only the solution of radius 5 is valid.

Edited on November 25, 2006, 2:23 pm

Posted by Dej Mar on 2006-11-25 09:20:05