The equation of a circle can be written as `(x-h)^2+(y-k)^2=r^2` where the center of the circle is at (h,k) on the cartesian coordinate plane, and the length of the radius is r.

We are given `x^2+y^2-6x+10y+9=0`

Rewrite as `x^2-6x+y^2+10y=-9`

Then use completing the square on the terms involving x, and...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The equation of a circle can be written as `(x-h)^2+(y-k)^2=r^2` where the center of the circle is at (h,k) on the cartesian coordinate plane, and the length of the radius is r.

We are given `x^2+y^2-6x+10y+9=0`

Rewrite as `x^2-6x+y^2+10y=-9`

Then use completing the square on the terms involving x, and the terms involving y:

`x^2-6x+9+y^2+10y+25=-9+9+25`

`(x-3)^2+(y+5)^2=25`

---------------------------------------------------------------

**The center is at (3,-5) and the radius is 5**

---------------------------------------------------------------