# How do you write an equation for a circle with center (-8, -5) and tangent (touching at 1 point) to the y-axis?

##### 1 Answer
Nov 25, 2016

Please see the explanation.

#### Explanation:

The standard form of the equation of a circle is:

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

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center and r is the radius.

Substitute the given center into the equation:

${\left(x - - 8\right)}^{2} + {\left(y - - 5\right)}^{2} = {r}^{2}$

Because the radius drawn to the point of tangency is always perpendicular, the point of tangency with the y axis must be $\left(0 , - 5\right)$

Substitute this point into the circle:

${\left(x - - 8\right)}^{2} + {\left(- 5 - - 5\right)}^{2} = {r}^{2}$

$r = 8$

The equation for the circle is:

${\left(x - - 8\right)}^{2} + {\left(y - - 5\right)}^{2} = {8}^{2}$