# How do you write the standard form of the equation of the circle that is tangent to the x-axis and has its center at (3,-2)?

Jul 12, 2016

${x}^{2} + {y}^{2} - 6 x + 4 y + 9 = 0$

#### Explanation:

The radius to the point of contact of the tangent (x-axis) is parallel to

the y-axis. So, the radius = |y-coordinate of the center| = 2.

And so, the equation is ${\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = {2}^{2}$. Rearranging,

${x}^{2} + {y}^{2} - 6 x + 4 y + 9 = 0$