# How do you find the equation of the circle passing through p1(-3,6)p2(-5,2),p3(3,-6)?

May 26, 2016

${\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(5 \sqrt{2}\right)}^{2}$

#### Explanation:

The circle equation is
$f \left(x , y\right) = {\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} - {r}^{2}$ in which ${x}_{0} , {y}_{0}$ are the circle coordinates and $r$ the radius.

If the points ${p}_{1} = \left({x}_{1} , {y}_{1}\right) , {p}_{2} = \left({x}_{2} , {y}_{2}\right) , {p}_{3} = \left({x}_{3} , {y}_{3}\right)$
obey the circle equation then

$f \left({x}_{1} , {y}_{1}\right) = f \left({x}_{2} , {y}_{2}\right) = f \left({x}_{3} , {y}_{3}\right) = 0$

or
$\left\{\begin{matrix}{\left({x}_{1} - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} - {r}^{2} = 0 \\ {\left({x}_{2} - {x}_{0}\right)}^{2} + {\left({y}_{2} - {y}_{0}\right)}^{2} - {r}^{2} = 0 \\ {\left({x}_{3} - {x}_{0}\right)}^{2} + {\left({y}_{3} - {y}_{0}\right)}^{2} - {r}^{2} = 0\end{matrix}\right.$

Subtracting the second from the first and the third from the second we have

$\left\{\begin{matrix}2 \left({x}_{2} - {x}_{1}\right) {x}_{0} + 2 \left({y}_{2} - {y}_{1}\right) {y}_{0} + \left({x}_{1}^{2} - {x}_{2}^{2} + {y}_{1}^{2} - {y}_{2}^{2}\right) = 0 \\ 2 \left({x}_{3} - {x}_{2}\right) {x}_{0} + 2 \left({y}_{3} - {y}_{2}\right) {y}_{0} + \left({x}_{2}^{2} - {x}_{3}^{2} + {y}_{2}^{2} - {y}_{3}^{2}\right) = 0\end{matrix}\right.$

Solving for ${x}_{0} , {y}_{0}$ after substituting the values for ${x}_{1} , {y}_{1} , {x}_{2} , {x}_{2} , {x}_{3} , {y}_{3}$ we get
${x}_{0} = 2 , {y}_{0} = 1$.

The radius is found by solving for $r$
${\left({x}_{1} - {x}_{0}\right)}^{2} + {\left({y}_{1} - {y}_{0}\right)}^{2} - {r}^{2} = 0$ so $r = 5 \sqrt{2}$