# How do you graph X^2+Y^2-16X+4Y+52=0?

Jan 14, 2017

This is a circle of radius $4$ with centre $\left(8 , - 2\right)$

#### Explanation:

Complete the square for both $x$ and $y$ in order to get this in the form of the standard equation of a circle:

Given:

${x}^{2} + {y}^{2} - 16 x + 4 y + 52 = 0$

Reorganise as:

$\textcolor{b l u e}{{x}^{2} - 16 x + 64} + \textcolor{g r e e n}{{y}^{2} + 4 y + 4} - 16 = 0$

That is:

$\textcolor{b l u e}{{x}^{2} - 2 \left(8 x\right) + {8}^{2}} + \textcolor{g r e e n}{{y}^{2} + 2 \left(2 y\right) + {2}^{2}} - {4}^{2} = 0$

Hence:

$\textcolor{b l u e}{{\left(x - 8\right)}^{2}} + \textcolor{g r e e n}{{\left(y + 2\right)}^{2}} = {4}^{2}$

which is (more or less) in the form:

$\textcolor{b l u e}{{\left(x - h\right)}^{2}} + \textcolor{g r e e n}{{\left(y - k\right)}^{2}} = {r}^{2}$

with $\left(h , k\right) = \left(8 , - 2\right)$ being the centre of the circle and $r = 4$ being the radius.

graph{(x^2+y^2-16x+4y+52)((x-8)^2+(y+2)^2-0.038)=0 [-4.08, 15.92, -6.84, 3.16]}