# How do you write the equation of the circle with a diameter that has endpoints at (8, 7) and (–4, –3)?

Jun 2, 2016

Find the center and radius of the circle, and then write it in standard form as

${\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = 61$

#### Explanation:

The standard equation of a circle centered at $\left({x}_{0} , {y}_{0}\right)$ with radius $r$ is ${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$. Thus, if we can find the center and the radius, we can write the equation.

As the midpoint of any diameter of a circle is the center of the circle, we can find the center by locating the midpoint of the line segment with endpoints $\left(8 , 7\right)$ and $\left(- 4 , - 3\right)$. The midpoint of between any two points $\left({a}_{1} , {b}_{1}\right)$ and $\left({a}_{2} , {b}_{2}\right)$ is $\left(\frac{{a}_{1} + {a}_{2}}{2} , \frac{{b}_{1} + {b}_{2}}{2}\right)$. Then, substituting our values, we have the midpoint, and thus the circle's center, at

$\left(\frac{8 - 4}{2} , \frac{7 - 3}{2}\right) = \left(2 , 2\right)$

Next, to find the radius of the circle, we can just calculate the distance from the center to either of the given points on the circle. The distance between two points $\left({a}_{1} , {b}_{1}\right) , \left({a}_{2} , {b}_{2}\right)$ is $\sqrt{{\left({a}_{2} - {a}_{1}\right)}^{2} + {\left({b}_{2} - {b}_{1}\right)}^{2}}$. Then, using the center at the point $\left(8 , 7\right)$, we have the radius as

$\sqrt{{\left(8 - 2\right)}^{2} + {\left(7 - 2\right)}^{2}} = \sqrt{36 + 25} = \sqrt{61}$

With both the center at the radius, we can now write the circle's equation:

${\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\sqrt{61}\right)}^{2}$

$\therefore {\left(x - 2\right)}^{2} + {\left(y - 2\right)}^{2} = 61$