How do you graph $x^{2} + y^{2} - 4 x - 2 y - 4 = 0$ $x^2+y^2-4x-2y-4=0$ ?

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Answer 1 · 2016-02-02T09:28:15

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

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:

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

We start from the equation given:

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

Now we can group terms with the same variable:

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

Now we can complete the squares of variables:

$x^{2} - 4 x + 4 + y^{2} - 2 y + 1 - 5 - 4 = 0$ $x^2-4x color(red)(+4) +y^2-2y color(red)(+1) color(red)(-5) -4=0$

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

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

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

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

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

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

How do you graph x^2+y^2-4x-2y-4=0x2+y2−4x−2y−4=0x^2+y^2-4x-2y-4=0?