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

Jul 28, 2016

This is a circle with centre $\left(1 , 2\right)$ and radius $5$.

#### Explanation:

Complete the square for both $x$ and $y$ ...

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

$= \textcolor{b l u e}{{x}^{2} - 2 x + 1} + \textcolor{g r e e n}{{y}^{2} - 4 y + 4} - 25$

$= {\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} - {5}^{2}$

Add ${5}^{2}$ to both ends and transpose to get:

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = {5}^{2}$

This is in the form:

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

the equation of a circle with centre $\left(h , k\right) = \left(1 , 2\right)$ and radius $r = 5$

graph{((x-1)^2+(y-2)^2-25)((x-1)^2+(y-2)^2-0.01) = 0 [-9.04, 10.96, -2.76, 7.24]}