Question

What are the center and radius of the circle defined by the equation `x^2+y^2-6x+4y+4=0`?

Answer 1

The center is `=(3,-2)` and the radius is `=3`

Explanation:

The standard equation of a circle is

`(x-a)^2+(y-b)^2=r^2`

Where the center is `=(a,b)` and the radius is `=r`

Here, we have

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

Transform this equation to the standar form by completing the square

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

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

Factorise

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

The center is `=(3,-2)` and the radius is `=3`

graph{(x^2+y^2-6x+4y+4)=0 [-14.08, 8.11, -5.51, 5.58]}

Answer 2

Center is at `(3,-2)`, radius is `3` units

Explanation:

Given: `x^2+y^2-6x+4y+4=0`.

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

Complete the square by adding `9` to both sides.

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

`=>(x-3)^2+(y+2)^2=9`

`=>(x-3)^2+(y+2)^2=3^2`

Now, compare it with the equation of a circle, given by:

`(x-a)^2+(y-b)^2=r^2`

where:

• `(a,b)` are the coordinates of the circle's center

• `r` is the radius of the circle

So, here the circle has a radius of `3` units, and has its center located at `(3,-2)`.

Graph of the circle:

graph{(x-3)^2+(y+2)^2=9 [-10, 10, -5, 5]}