# How do you graph x^2 + y^2 - 8x - 10y + 5=0?

Feb 21, 2016

To graph the equation x^2+y^2−8x−10y+5=0, draw a circle with center at $\left(4 , 5\right)$ and radius $6$.

#### Explanation:

As the equation x^2+y^2−8x−10y+5=0 is of degree $2$ and coefficients of ${x}^{2}$ and ${y}^{2}$ are equal and that of $x y$ is $0$, it is clearly an equation of a circle.

Let us convert it into sum of squares as follows

(x^2−8x+16)+(y^2−10y+25)=16+25-5 or

${\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} = 36$ or

${\left(x - 4\right)}^{2} + {\left(y - 5\right)}^{2} = {6}^{2}$

which is the equation of a circle with center at $\left(4 , 5\right)$ and radius $6$.

Hence, to graph the equation x^2+y^2−8x−10y+5=0, draw a circle with center at $\left(4 , 5\right)$ and radius $6$.

graph{(x^2+y^2-8x-10y+5)(x^2+y^2-8x-10y+40.95)=0 [-10, 18, -2, 12]}