# How do you find general form of circle tangent to the y-axis and has a center of (-6, 7)?

Dec 31, 2015

Use a little reasoning to find:

${\left(x + 6\right)}^{2} + {\left(y - 7\right)}^{2} = {6}^{2}$

or if really fussy:

${\left(x - \left(- 6\right)\right)}^{2} + {\left(y - 7\right)}^{2} = {6}^{2}$

#### Explanation:

The general equation of a circle with centre $\left(h , k\right)$ and radius $r$ is:

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

In our case $\left(h , k\right) = \left(- 6 , 7\right)$ and the radius $r$ must be $6$ for the circle to touch the $y$ axis.

graph{((x-(-6))^2+(y-7)^2-6^2)((x-(-6))^2+(y-7)^2-0.012)=0 [-24.75, 15.25, -3.52, 16.48]}