Question

A circle has a center of (5,-5) and goes through (6,-2). What is the radius?

Answer 1

Start with the equation of a circle whose center is at (h,k) and whose radius is r:

(x-h)^2 + (y-k)^2 = r^2

Substituting the given coordinates of the center:

(x-5)^2 + (y-[-5])^2 = r^2, or (x-5)^2 + (y+5)^2 = r^2

Substituding the given coordinates of a point (6,-2) on the circle:

(6-5)^2 + (-2+5)^2 = r^2

Simplifying:

1^2 + 3^2 = r^2, or 1 + 9 = r^2, or 10= r^2.  Then r = sqrt(10).