Question

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)?

Answer 1

`x^2+y^2-6x+4y+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 `(x-3)^2+(y+2)^2=2^2`. Rearranging,

`x^2+y^2-6x+4y+9=0`