Expert: Socrates - 9/30/2009

Question
32. For question 32, find the standard equation of the circle satisfying the given conditions. A diameter has endpoints (6,1) and (-2,3).

Answer
For the standard equation of a circle you need to find the center and radius

The midpoint of a diameter will always be the center of the circle.

The midpoint of the segment joining (6,1) and (-2,3) will be ((6-2)/2 , (1+3)/2) = (2,2)

The center of the circle is (2,2)

The radius is one half the length of a diameter. The length of the segment joining (6,1) and
(-2,3) will be √( (6 + 2)^2 + (1 - 3)^2 ) = √( (8)^2 + (-2)^2 ) = √(64 + 4) = √68

The radius is then √68 / 2

The standard equation for a circle with center (h,k) and radius r is

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

Substitute h = 2  , k = 2 , r = √68 / 2 and get the equation of the circle


(x-2)^2 + (y-2)^2 = 17