# How do you find the equation of a circle that passes through the following points: (5,3) (-2,2) (-1,-5)?

Jul 28, 2016

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

#### Explanation:

Notice that:

$\left(- 2 , 2\right) - \left(5 , 3\right) = \left(- 7 , - 1\right)$

$\left(- 1 , - 5\right) - \left(- 2 , 2\right) = \left(1 , - 7\right)$

So the line segment joining $\left(5 , 3\right)$ and $\left(- 2 , 2\right)$ is orthogonal to and the same length as the line segment joining $\left(- 2 , 2\right)$ and $\left(- 1 , - 5\right)$.

Thus these three points form an isosceles triangle with angles $\frac{\pi}{4}$, $\frac{\pi}{4}$ and $\frac{\pi}{2}$.

The centre of the circle passing through them is therefore the midpoint of the line segment joining $\left(5 , 3\right)$ and $\left(- 1 , - 5\right)$, namely:

$\left(\frac{5 - 1}{2} , \frac{3 - 5}{2}\right) = \left(2 , - 1\right)$

The radius is the distance from this point to any of the original three points, say $\left(5 , 3\right)$:

$r = \sqrt{{\left(5 - 2\right)}^{2} + {\left(3 - \left(- 1\right)\right)}^{2}} = \sqrt{{3}^{2} + {4}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

The equation of the circle can therefore be written:

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

graph{((x-2)^2+(y+1)^2-25)((x-2)^2+(y+1)^2-0.02)((x+2)^2+(y-2)^2-0.04)((x-5)^2+(y-3)^2-0.04)((x+1)^2+(y+5)^2-0.04) = 0 [-9.29, 10.71, -5.96, 4.04]}