Question

Find an equation of the circle with center at (-9,2) and passing through (2,-2) in the form of (x-A)^2+(y-B)^2=C where A,B,C are constant.

Answer 1

The center going through (-9,2) gives you A and B
(x - - 9)^2 + (y - 2)^2 = r^2
(x + 9)^2 + (y - 2)^2 = r^2

Going from the center to the given point gives you the radius. We need the r^2 which is the way the distance formula works: it will give you r^2. c = r^2

d^2 = (x2 - x1)^2 + (y2 - y1)^2
x2 = -9
x1 = 2
y2 = 2
y1 = -2
r^2 = (-9 - 2)^2 + (2 - - 2)^2
r^2 = (- 11 ) ^2 + (4)^2
r^2 = 121 + 16
r^2 o= 137
So we get
(x + 9)^2 + (y - 2)^2 = 137