# How do you find the center and radius of the circle x^2-8x+y^2-12y= -51?

Nov 1, 2016

The center is the point $\left(4 , 6\right)$ and the radius, $r = 1$

#### Explanation:

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

${x}^{2} - 8 x + {h}^{2} + {y}^{2} - 12 y + {k}^{2} = {h}^{2} + {k}^{2} - 51$

Use the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ to find the value of $h$ and ${h}^{2}$:

${x}^{2} - 2 h x + {h}^{2} = {x}^{2} - 8 x + {h}^{2}$

$- 2 h x = - 8 x$

$h = 4$

${h}^{2} = 16$

Write the left side as a perfect square and substitute 16 for ${h}^{2}$ on the right:

${\left(x - 4\right)}^{2} + {y}^{2} - 12 y + {k}^{2} = 16 + {k}^{2} - 51$

Use the pattern ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$ to find the value of $k$ and ${k}^{2}$:

${y}^{2} - 2 k y + {k}^{2} = {y}^{2} - 12 y + {k}^{2}$

$- 2 k y = - 12 y$

$k = 6$

${k}^{2} = 36$

Write the left side as a perfect square and substitute 36 for ${k}^{2}$ on the right:

${\left(x - 4\right)}^{2} + {\left(y - 6\right)}^{2} = 16 + 36 - 51$

${\left(x - 4\right)}^{2} + {\left(y - 6\right)}^{2} = 1$