Find the equation of the circle that passes through (2,2) and tangent to the line x=1 and x=6.

Answer 1

the two lines are 5 apart, so the radius is 5/2, with the center at (7/2,k) so

(x - 7/2)^ + (y-k)^2 = (5/2)^2

The point (2,2) is on the line, so

(2 - 7/2)^2 + (2-k)^2 = (5/2)^2
k = 0 or 4

(x - 7/2)^2 + (y-4)^2 = (5/2)^2
(x - 7/2)^2 + y^2 = (5/2)^2

So, the question is misleading. There are two circles that fit the conditions:

http://www.wolframalpha.com/input/?i=plot+%28x+-+7%2F2%29^2+%2B+%28y-4%29^2+%3D+%285%2F2%29^2%2C+%28x+-+7%2F2%29^2+%2B+y^2+%3D+%285%2F2%29^2+and+x%3D1+and+x%3D6