# How do you write an equation for a circle given center (0,3) and radius is 7 units?

Jan 10, 2017

${x}^{2} + {\left(y - 3\right)}^{2} = {7}^{2}$, or
${x}^{2} + {y}^{2} - 6 y - 40 = 0$

#### Explanation:

The Cartesian equation of a circle with centre $\left(a , b\right)$ and radius $r$ is:

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

So a circle with centre $\left(0 , 3\right)$ and radius $7$ would be:

${\left(x - 0\right)}^{2} + {\left(y - 3\right)}^{2} = {7}^{2}$
$\therefore \setminus \setminus \setminus \setminus \setminus {x}^{2} + {\left(y - 3\right)}^{2} = {7}^{2}$

We can multiply out if required to get:

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus {x}^{2} + {y}^{2} - 6 y + 9 = 49$
$\therefore {x}^{2} + {y}^{2} - 6 y - 40 = 0$