# How do you write the equation of a circle given endpoints of a diameter (-4,1) and (4,-5)?

Oct 6, 2016

${x}^{2} + {y}^{2} + 4 y - 21 = 0$

#### Explanation:

Strategy:

• The center of the circle will be the midpoint of the diameter;
• The length of the radius will be half the length of the diameter.

The coordinates of the midpoint of the diameter are given by the average of the coordinates of the endpoints, so

$C = \left({C}_{x} , {C}_{y}\right) = \left(\frac{- 4 + 4}{2} , \frac{1 - 5}{2}\right) = \left(0 , - 2\right)$

The length of the diameter is given by Pytagora's theorem, so we have $d = \sqrt{{d}_{x}^{2} + {d}_{y}^{2}} = 10$, so the radius has length $5$.

Now that we know everything we need, we can write the equation using the fact that a circle of radius $r$ and center coordinates $\left({C}_{x} , {C}_{y}\right)$ has equation

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

In our case,

${\left(x - 0\right)}^{2} + {\left(y + 2\right)}^{2} = 25$

Which we can expand as

${x}^{2} + {y}^{2} + 4 y + 4 = 25$

and thus

${x}^{2} + {y}^{2} + 4 y - 21 = 0$