# How do you write the equation of the circle where C(1,-3) and D(-3,7) are the endpoints of a diameter?

May 9, 2016

${\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 29$

#### Explanation:

The midpoint of the diameter $C D$ must be the center of the circle:
$\textcolor{w h i t e}{\text{XXX}}$center is at $\left({c}_{x} , {c}_{y}\right) = \left(\frac{1 + \left(- 3\right)}{2} , \frac{\left(- 3\right) + 7}{2}\right) = \left(- 1 , 2\right)$

The length of the diameter is given by the Pythagorean Theorem:
$\textcolor{w h i t e}{\text{XXX}} \left\mid C D \right\mid = \sqrt{{\left(1 - \left(- 3\right)\right)}^{2} + {\left(- 3 - 7\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} = \sqrt{116}$

$\textcolor{w h i t e}{\text{XXX}} = 2 \sqrt{29}$

which implies that the radius of the circle is
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{29}$

The general formula for a circle with center $\left({c}_{x} , {c}_{y}\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {c}_{x}\right)}^{2} + {\left(y - {c}_{y}\right)}^{2} = {r}^{2}$

In this case:
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 29$

For verification purposes here is the graph of this circle equation with the given diameter endpoints:
graph{((x+1)^2+(y-2)^2-29)((x-1)^2+(y+3)^2-0.02)((x+3)^2+(y-7)^2-0.02)=0 [-12.53, 12.78, -4.8, 7.86]}