Question

How do you find the center and radius of a circle using a polynomial `(x^2) + (y^2) + 6x - 4y = 12`?

Answer 1

Re-write the equation in standard form for a circle to get:
center: `(-3,2)` and radius: `5`

Explanation:

Given
`color(white)("XXX")x^2+y^2+6x-4y=12`

Complete the square for each of the `x` and `y` components:
`color(white)("XXX")(x^2+6xcolor(red)(+9))+(y^2-4ycolor(blue)(+4)) = 12 color(red)(+9)color(blue)(+4)`

Re-write as squared binomials and squared constant:
`color(white)("XXX")(xcolor(green)(+3))^2+(y-color(orange)(2))^2=color(cyan)(5)^2`

Since the standard form for a circle with center `(color(green)(a),color(orange)(b))` and radius `color(cyan)(r)` is
`color(white)("XXX")(x-color(green)(a))^2+(y-color(orange)(b))^2 = color(cyan)(r)^2`

The given circle has a center `(color(green)(-3),color(orange)(2))` and radius `color(cyan)(5)`
graph{x^2+y^2+6x-4y=12 [-11.955, 8.045, -3, 7]}