# How do you graph  x^2+y^2-4x+10y+20=0?

Jan 26, 2016

Graphs a circle with centre $\left(2 , - 5\right)$ and the radius $r = 3$

#### Explanation:

This equation can be recognised as the equation of a circle - see Conic sections for a description of the patterns of the conic equations.

We need to reorganise and simplify it in order to get it into the standard form for a circle that will give us the centre and radius, after which it will be very easy to graph.

${x}^{2} - 4 x + {y}^{2} + 10 y + 20 = 0$

Using completing the squares gives us
${\left(x - 2\right)}^{2} - 4 + {\left(y + 5\right)}^{2} - 25 + 20 = 0$

${\left(x - 2\right)}^{2} + {\left(y + 5\right)}^{2} = 9$

This is now in the form ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$ where $\left(h , k\right)$ is the centre and $r$ is the radius.
therefore the centre is $\left(2 , - 5\right)$ and the radius $r = 3$

It is now possible to graph the circle.