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

Feb 2, 2016

It is an equation of a circle with center in C=(2;1) and radius $r = 3$

#### Explanation:

From this form you can find out that the equation represents a circle.
To find its coordinates and radius you should transform it to form of:

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

We start from the equation given:

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

Now we can group terms with the same variable:

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

Now we can complete the squares of variables:

${x}^{2} - 4 x \textcolor{red}{+ 4} + {y}^{2} - 2 y \textcolor{red}{+ 1} \textcolor{red}{- 5} - 4 = 0$

I added $5$ so I had to substract 5 to keep the equation balanced.

Now we can write the expressions with $x$ and $y$ as squares:

${\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} - 9 = 0$

Finally I can move $9$ to right side to get the form (1)

${\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} = 9$

From this equation I can read that the equation shows a circle with center in C=(2;1) and radius $r = 3$