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

Nov 1, 2015

$0 = {x}^{2} + {y}^{2} - 8 x + 6 y + 16 = {\left(x - 4\right)}^{2} + {\left(y + 3\right)}^{2} - {3}^{2}$

is a circle of radius $3$ with centre $\left(4 , - 3\right)$

#### Explanation:

The equation of a circle of radius $r$ centred at $\left(a , b\right)$ can be written:

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

We are given:

$0 = {x}^{2} + {y}^{2} - 8 x + 6 y + 16$

$= {x}^{2} - 8 x + 16 + {y}^{2} + 6 y + 9 - 9$

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

So:

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

which is in the form of the equation of a circle of radius $3$ centre $\left(4 , - 3\right)$

graph{(x-4)^2+(y+3)^2 = 3^2 [-7.875, 12.125, -7.8, 2.2]}