SOLUTION: write the standard form of the equation of the circle that passes through the point at (2, -2) and has its center at (-2, 3).

The standard form of a circle is

     (x - h)² + (y - k)² = r²

where the center is (h,k) and the radius is r

We know the center (h,k) = (-2,3), so h = -2 and k = 3.

So we can fill those in

     (x - h)² + (y - k)² = r²

(x - (-2))² + (y - (3))² = r²

     (x + 2)² + (y - 3)² = r²

We plot the center and the point the circle goes through:



To find the radius, some people use the distance formula, but

you can just substitute the point (x,y) = (2,-2) into the equation

     (x + 2)² + (y - 3)² = r²
    (2 + 2)² + (-2 - 3)² = r²
              4² + (-5)² = r²
                 16 + 25 = r²
                      41 = r²

Of course the radius is  but r² = 41 is what we need
to finish the standard form: 

     (x + 2)² + (y - 3)² = 41

Edwin