College Algebra

Question

Find an equation of the circle that has center (2,-1) and passes through (6,-5).

Answer 1

(x-2)^2 + (y+1)^2 = r^2

plug in (6,-5) to find r

(6-2)^2 + (-5+1)^2 = r^2

4^2 + (-4)^2 = r^2

16 +16 = r^2

r^2 = 32

r = sqr32 = 4sqr2

(x-2)^2 + (y+1)^2 = 32 is the circle's equation with given center through the given point