# How do you find the equation of a circle with diameter has endpoints (-2, 3) and (4, -1)?

Aug 14, 2016

${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = 13$

#### Explanation:

The radius of the circle is half of the length of the diameter, so half the distance between these two endpoints:

$\frac{1}{2} \sqrt{{\left(4 - \left(- 2\right)\right)}^{2} + {\left(\left(- 1\right) - 3\right)}^{2}} = \frac{1}{2} \sqrt{36 + 16} = \frac{1}{2} \sqrt{52} = \sqrt{13}$

The centre of the circle is the midpoint of the diameter:

$\left(\frac{\left(- 2\right) + 4}{2} , \frac{3 + \left(- 1\right)}{2}\right) = \left(1 , 1\right)$

The equation of a circle with centre $\left(h , k\right)$ and radius $r$ may be written:

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

So in our case:

${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = 13$

graph{((x-1)^2+(y-1)^2-13)((x-1)^2+(y-1)^2-0.02)((x+2)^2+(y-3)^2-0.02)((x-4)^2+(y+1)^2-0.02) = 0 [-10, 10, -5, 5]}