Question

How do you write the equation of the circle in standard form, identify the center and radius of `9x^2+9y^2+54x-36y+17=0`?

Answer 1

Equation of the circle in standard form is `(x-(-3))^2+(y-2)^2=(10/3)^2` and its center is `(-3,2)` and radius is `10/3`

Explanation:

Standard form of equation of a circle is

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

which has a radius `r` and its center is `(h,k)`

Now `9x^2+9y^2+54x-36y+17=0` can be written as (dividing each term by `9`

`x^2+y^2+6x-4y=-17/9`

or `x^2+6x+9+y^2-4y+4=-17/9+9+4`

or `(x+3)^2+(y-2)^2=100/9`

or `(x-(-3))^2+(y-2)^2=(10/3)^2`

and as is observed. its center is `(-3,2)` and radius is `10/3`
graph{9x^2+9y^2+54x-36y+17=0 [-13.125, 6.875, -3.32, 6.68]}