# How do you find the general form of the equation of a circle given the center at the point (-3, 1) and tangent to the y-axis?

Oct 17, 2016

Please see the explanation for the process. The equation of the circle is:

${\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {3}^{2}$

#### Explanation:

The general equation for a circle is:

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

where $\left(h , k\right)$ is the center point and $r$ is the radius.

The center is $\left(- 3 , 1\right)$; this gives us the equation:

${\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {r}^{2}$

We are told that the y axis is tangent, therefore, the x coordinate is 0 and the y coordinate is the same as the center, 1. To find the value of r, temporarily substitute the point $\left(0 , 1\right)$ for x and y in the equation:

${\left(0 - - 3\right)}^{2} + {\left(1 - 1\right)}^{2} = {r}^{2}$

$r = 3$

The equation of the circle is:

${\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {3}^{2}$