# How do you write the equation in standard form of the circle for; endpoints of the diameter A(14, -7) and B(-10,9)?

Jan 12, 2016

${\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} = 208$

#### Explanation:

The general form of a circle centred at $\left(a , b\right)$ and with radius $r$ is ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$.

So we first need to find the radius and the centre before we can find the equation.

Length of diameter will be distance between its endpoints.

$\therefore d = \sqrt{{\left(14 - \left(- 10\right)\right)}^{2} + {\left(- 7 - 9\right)}^{2}} = \sqrt{832}$.

Therefore the radius $r = \frac{1}{2} d = \frac{1}{2} \sqrt{832}$.

This implies that ${r}^{2} = \frac{832}{4} = 208$.

The centre of the circle will be the midpoint of the diameter, ie

(a,b)=((14-10)/2;(9-7)/2)=(2,1).

Therefore the standard equation of this circle will be

${\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} = 208$.