# How do you write the equation of the circle with center at (–6, 10) and diameter = 6?

Sep 9, 2016

${\left(x + 6\right)}^{2} + {\left(y - 10\right)}^{2} = 9$

#### Explanation:

The equation of a circle of radius $\left(r\right)$ centred at the point $\left({x}_{1} , {y}_{1}\right)$ is:

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

In this example: ${x}_{1} = - 6 ,$ ${y}_{1} = 10$ and $r = \frac{6}{2} = 3$

Hence the equation of this circle is:
${\left(x + 6\right)}^{2} + {\left(y - 10\right)}^{2} = {3}^{2}$
$\to {\left(x + 6\right)}^{2} + {\left(y - 10\right)}^{2} = 9$