# How do you write the standard form of the equation of the circle with Center: (3, -2); Radius: 3?

Dec 13, 2015

The standard form of the equation of the circle would be ${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = 9$.

#### Explanation:

The standard form of the equation of a circle is:

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

$x$ and $y$ are the $x$ and $y$ variables, ${x}_{1}$ is the x-coordinate of the center, ${y}_{1}$ is the y-coordinate of the center, and $r$ is the radius of the circle.

In order to place the center of the circle at point (3, -2), simply replace ${x}_{1}$ with 3 and ${y}_{1}$ with -2.

The equation is now:

${\left(x - 3\right)}^{2} + {\left(y - \left(- 2\right)\right)}^{2} = {r}^{2}$

This can be simplified as:

${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = {r}^{2}$

Finally, replace $r$ with the radius of the circle.

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

The final equation is:

${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = 9$