Let the equation be x^2 + y^2 + Ax + By + C = 0x2+y2+Ax+By+C=0
Accordingly, we can write a system of equations.
Equation #1:
(-9)^2 + (-16)^2 + A(-9) + B(-16) + C = 0(−9)2+(−16)2+A(−9)+B(−16)+C=0
81 + 256 - 9A - 16B + C = 081+256−9A−16B+C=0
337 - 9A - 16B + C = 0337−9A−16B+C=0
Equation #2
(-9)^2 + (32)^2 - 9A + 32B + C = 0(−9)2+(32)2−9A+32B+C=0
81 + 1024 - 9A + 32B + C = 081+1024−9A+32B+C=0
1105 - 9A + 32B + C = 01105−9A+32B+C=0
Equation #3
(22)^2 + (15)^2 + 22a + 15B + C = 0(22)2+(15)2+22a+15B+C=0
709 + 22A + 15A + C = 0709+22A+15A+C=0
The system therefore is {(337 - 9A - 16B + C = 0),(1105 - 9A + 32B + C = 0), (709 + 22A + 15B + C = 0):}
After solving, either using algebra, a C.A.S (computer algebra system) or matrices, you should get solutions of A = 4, B = -16, C = -557.
Hence, the equation of the circle is x^2 + y^2 + 4x - 16y -557 = 0.
Hopefully this helps!