Question

How do you find the center and radius of the circle given `x^2-12x+84=-y^2+16y`?

Answer 1

Center (6, 8) and radius = 4

Explanation:

rewrite equation as
x^2 - 12x +y^2-16y +84 =0
(x^2 -12x +36) + (y^2-16y +64) +84 -36-64 =0
(x-6)^2 +(y-8)^2 -16 =0
(x-6)^2 +(y-8)^2 = 16 = 4^2
So center is (6,8) and radius 4

Answer 2

Write the equation as: `(x - 6)^2 + (y - 8)^2 = 4^2`
Center: `(6,8)`
Radius: 4

Explanation:

You need to put the equation into standard form:

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

because we know that `(h, k)` is the center and `r` is the radius.

In the given equation, move the constant term to the right and all of the other terms to the left:

`x^2 - 12x + y^2 - 16y = -84`

Add `h^2 and k^2` to both sides:

`x^2 - 12x + h^2 + y^2 - 16y + k^2= -84 +h^2 + k^2`

Because `(x - h)^2 = x^2 - 2hx + h^2`, we can set middle term of this pattern equal to the middle term of the given equation and then find the value of `h` and `h^2`:

`-2hx = -12x`

`h = 6, h^2 = 36`

Substitute 36 for every `h^2`

`x^2 - 12x + 36 + y^2 - 16y + k^2= -84 +36 + k^2`

We know that the left side is a perfect square and the right side simplifies a bit:

`(x - 6)^2 + y^2 - 16y + k^2= -48 + k^2`

We do the same thing for the middle term to find `k` and `k^2`:

`-2ky = -16y`

`k = 8, k^2 = 64`

Substitute 64 for every `k^2`:

`(x - 6)^2 + y^2 - 16y + 64= -48 + 64`

We know that the left is a perfect square and the right side simplifies:

`(x - 6)^2 + (y - 8)^2 = 16`

Make the radius obvious by writing the ride side as a square:

`(x - 6)^2 + (y - 8)^2 = 4^2`