Question

How do you find the center and radius of the circle given `x^2+y^2-18x-18y+53=0`?

Answer 1

Complete the square for the x and y terms to find the center `=(9,9)` and the radius `r=sqrt109`.

Explanation:

`x^2+y^2-18x-18y+53=0`

Use a method called "Completing the Square".

1) Group the x terms and y terms. Move the constant term to the right side of the equation by subtracting 53 from both sides.

`(x^2color(limegreen)(-18)xcolor(white)(aaa))+(y^2color(magenta)(-18)ycolor(white)(aaa))=-53`

2) Divide the coefficient of the x term by 2 and then square it.

`color(limegreen)(-18)/2= -9color(white)(aaa)(-9)^2 =color(red)(81)`

3) Add the result to both sides.

`(x^2color(limegreen)(-18)x+color(red)(81))+(y^2color(magenta)(-18)ycolor(white)(aaa))=-53+color(red)(81)`

4) Divide the coefficient of the y term by 2 and then square it.

`color(magenta)(-18)/2=-9color(white)(aaa)(-9)^2=color(blue)(81)`

5) Add the result to both sides.

`(x^2-18x+color(red)(81))+(y^2-18y+color(blue)(81))=-53+color(red)(81)+color(blue)(81)`

Factor each set of parentheses. Note that the `-9` in factored form is the same number you got when dividing the coefficient of the middle term.

`(x-9)(x-9)+(y-9)(y-9)=109`

Rewrite as the square of a binomial.

`(x-9)^2+(y-9)^2=109`

Compare this equation to the equation of a circle
`(x-h)^2+(y-k)^2=r^2`
where `(h,k)` is the center and `r` is the radius.

In this example, the center `(h,k)=(9,9)` and the radius `r=sqrt109`