# What is the equation of a circle with center (2, -3) and radius of 3?

Jul 25, 2016

${x}^{2} + {y}^{2} - 4 x + 6 y + 4 = 0$.

#### Explanation:

The eqn. of a circle with centre $\left(h , k\right)$ and radius $r$ is given by,

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

Accordingly, the reqd. eqn. is,

${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = {3}^{2}$, i.e.,

${x}^{2} + {y}^{2} - 4 x + 6 y + 4 = 0$.

Hi.
Equation of standard circle i.e circle having its centre at origin is
${x}^{2}$+${y}^{2}$=${r}^{2}$
${\left(x - X 1\right)}^{2}$+${\left(y - Y 1\right)}^{2}$=${r}^{2}$
${\left(x - 2\right)}^{2}$+${\left(y - \left(- 3\right)\right)}^{2}$=9