# How do you graph x^2+y^2+6x-16y+48=0?

##### 1 Answer
Feb 1, 2016

Rearrange and group the terms to get the expression into the standard form of one of the conics - in this case a circle. You can then identify the key points that you need to be able to graph it.

#### Explanation:

${x}^{2} + 6 x + {y}^{2} - 16 y + 48 = 0$

${\left(x + 3\right)}^{2} - 9 + {\left(y - 8\right)}^{2} - 64 + 48 = 0$

${\left(x + 3\right)}^{2} + {\left(y - 8\right)}^{2} = 73 - 48 = 25$

This is now recognisably a circle in the form

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

where $\left(h , k\right)$ is the centre of the circle and $r$ is the radius.

In this example $\left(- 3 , 8\right)$ is the centre and the radius is $5$