# How do you graph x^2+y^2-8x+6y+16=0?

Nov 15, 2016

Set your compass to a radius of 3, put the centerpoint at $\left(4 , - 3\right)$, and draw a circle.

#### Explanation:

This is a circle. To find the center and the radius, we must complete the squares for both x terms and y terms:

Add ${h}^{2} + {k}^{2} - 16$ to both sides:

${x}^{2} - 8 x + {h}^{2} + {y}^{2} + 6 y + {k}^{2} = {h}^{2} + {k}^{2} - 16$

Use the linear x term to find the value of h by setting it equal to -2xh:

$- 2 h x = - 8 x$

$h = 4$

Substitute ${\left(x - 4\right)}^{2}$ for the x terms on the left and 16 for ${h}^{2}$ on the right:

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

Use the linear y term to find the value of k by setting it equal to -2yk:

$- 2 y k = 6 y$

$k = - 3$

Substitute ${\left(y - - 3\right)}^{2}$ for the y terms on the left and 9 for ${k}^{2}$ on the right:

${\left(x - 4\right)}^{2} + {\left(y - - 3\right)}^{2} = 16 + 9 - 16 = 9 = {3}^{2}$

The final equation is:

${\left(x - 4\right)}^{2} + {\left(y - - 3\right)}^{2} = {3}^{2}$

This is a circle with a center of $\left(4 , - 3\right)$ and a radius of 3.