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

Oct 13, 2016

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

Oct 13, 2016

Write the equation as: ${\left(x - 6\right)}^{2} + {\left(y - 8\right)}^{2} = {4}^{2}$
Center: $\left(6 , 8\right)$

#### Explanation:

You need to put the equation into standard form:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{=} {r}^{2}$'

because we know that $\left(h , k\right)$ 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:

${x}^{2} - 12 x + {y}^{2} - 16 y = - 84$

Add ${h}^{2} \mathmr{and} {k}^{2}$ to both sides:

${x}^{2} - 12 x + {h}^{2} + {y}^{2} - 16 y + {k}^{2} = - 84 + {h}^{2} + {k}^{2}$

Because ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, 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 ${h}^{2}$:

$- 2 h x = - 12 x$

$h = 6 , {h}^{2} = 36$

Substitute 36 for every ${h}^{2}$

${x}^{2} - 12 x + 36 + {y}^{2} - 16 y + {k}^{2} = - 84 + 36 + {k}^{2}$

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

${\left(x - 6\right)}^{2} + {y}^{2} - 16 y + {k}^{2} = - 48 + {k}^{2}$

We do the same thing for the middle term to find $k$ and ${k}^{2}$:

$- 2 k y = - 16 y$

$k = 8 , {k}^{2} = 64$

Substitute 64 for every ${k}^{2}$:

${\left(x - 6\right)}^{2} + {y}^{2} - 16 y + 64 = - 48 + 64$

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

${\left(x - 6\right)}^{2} + {\left(y - 8\right)}^{2} = 16$

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

${\left(x - 6\right)}^{2} + {\left(y - 8\right)}^{2} = {4}^{2}$