Question

How do you find the center and radius of the circle `9x^2 + 54x + 9y^2 − 6y + 55 = 0`?

Answer 1

center `(-3, 1/3)`

Radius `sqrt3`

Explanation:

Given -

`9x^2+54x+9y^2-6y+55=0`

Let us rewrite the equation in the standard form -

`9(x^2+6x)+9(y^2-2/3y)=-55`

`9(x^2+6x+9)+9(y^2-2/3y+1/9)=-55+81+1`

`9(x^2+6x+9)+9(y^2-2/3y+1/9)=27`

`(x^2+6x+9)+(y^2-2/3y+1/9)=27/9=3`

`(x+3)^2+(y-1/3)^2=3`

Now it is in the form of -

(x-h)^2+(y-k)^2=a^2

In this case `(h, k)` is the center and `a` is the radius

Apply this to our case

center `(-3, 1/3)`

Radius `sqrt3`