Question

Write the equation of the circle in standard form. Then find the center x2 + y2 - 22x - 12y + 121 = 0

Answer 1

Standard form:

`(x-11)^2+(y-6)^2=36`

or

`(x-11)^2+(y-6)^2=6^2`

The center is
`(11,6)` and the radius is
`6`.

Explanation:

`x^2+y^2-22x-12y+121=0`

We will group terms with
`x` together and also group terms with
`y` together.

`x^2-22x+y^2-12y+121=0`

We will not subtract 121 on both sides.

`x^2-22x+y^2-12y=-121`

We are about to complete the square both both the
`x` terms and then the
`y` terms.

Whatever we add on one side, we must add to the other.

`(x-11)^2+(y-6)^2=-121+121+36`

`(x-11)^2+(y-6)^2=0+36`

`(x-11)^2+(y-6)^2=36`

We can also write it as:

`(x-11)^2+(y-6)^2=6^2`

Now it it easy to compare to:

`(x-h)^2+(y-k)^2=r^2`

to find the center
`(h,k)` and the radius,
`r`.

The center is
`(11,6)` and the radius is
`6`.