# What is the standard form of the equation of a circle passes through the points (–9, –16), (–9, 32), and (22, 15)?

Nov 30, 2016

Let the equation be ${x}^{2} + {y}^{2} + A x + B y + C = 0$

Accordingly, we can write a system of equations.

Equation 1:

${\left(- 9\right)}^{2} + {\left(- 16\right)}^{2} + A \left(- 9\right) + B \left(- 16\right) + C = 0$

$81 + 256 - 9 A - 16 B + C = 0$

$337 - 9 A - 16 B + C = 0$

Equation 2

${\left(- 9\right)}^{2} + {\left(32\right)}^{2} - 9 A + 32 B + C = 0$

$81 + 1024 - 9 A + 32 B + C = 0$

$1105 - 9 A + 32 B + C = 0$

Equation #3

${\left(22\right)}^{2} + {\left(15\right)}^{2} + 22 a + 15 B + C = 0$

$709 + 22 A + 15 A + C = 0$

The system therefore is $\left\{\begin{matrix}337 - 9 A - 16 B + C = 0 \\ 1105 - 9 A + 32 B + C = 0 \\ 709 + 22 A + 15 B + C = 0\end{matrix}\right.$

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 ${x}^{2} + {y}^{2} + 4 x - 16 y - 557 = 0$.

Hopefully this helps!