# How do you write the equation of the circle whose diameter goes from A(7,-2) to B(1,12)?

Jan 3, 2016

${\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} = 58$

#### Explanation:

The general equation for a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where $\left(h , k\right)$ is the center of the circle and $r$ is the radius. So, in order to find the equation of the circle, we need to know both its center and its radius length.

The center of the circle will be the midpoint of the endpoints of the diameter. The midpoint formula is as follows:

$\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

Hence the midpoint:

$\left(\frac{7 + 1}{2} , \frac{- 2 + 12}{2}\right) = \left(4 , 5\right)$

The remaining piece of information we need is the radius, which can be found through finding the distance from the center point to either one of the diameter's endpoints, which lie on the circle. First, the distance formula:

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

I'll choose $\left(1 , 12\right)$, along with the center at $\left(4 , 5\right)$.

$d = \sqrt{{\left(4 - 1\right)}^{2} + {\left(12 - 5\right)}^{2}}$
$= \sqrt{9 + 49}$
$= \sqrt{58}$

The radius has length $\sqrt{58}$.

Relate the center $\left(4 , 5\right)$ and radius $\sqrt{58}$ in the equation of a circle.

${\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} = {\left(\sqrt{58}\right)}^{2}$

$\implies {\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} = 58$

graph{(x-4)^2+(y-5)^2=58 [-13.06, 22.99, -3.42, 14.6]}