# How do you graph x^2 + y^2 – 4x + 22y + 61 = 0?

Mar 11, 2016

This is a circle, centre $\left(2 , - 11\right)$, with radius $8$

#### Explanation:

$0 = {x}^{2} + {y}^{2} - 4 x + 22 y + 61$

$= {x}^{2} - 4 x + 4 + {y}^{2} + 22 y + 121 - 64$

$= {\left(x - 2\right)}^{2} + {\left(y + 11\right)}^{2} - {8}^{2}$

Add ${8}^{2}$ to both ends and transpose to get:

${\left(x - 2\right)}^{2} + {\left(y + 11\right)}^{2} = {8}^{2}$

which is (almost) in the standard form:

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

the equation of a circle with centre $\left(h , k\right) = \left(2 , - 11\right)$ of radius $r = 8$.

graph{(x^2+y^2-4x+22y+61)((x-2)^2+(y+11)^2-0.06) = 0 [-17.8, 22.2, -19.96, 0.04]}