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

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Answer 1 · 2018-07-17T12:28:19

Please see the explanation below

The equation is

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

Complete the square

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

$x^{2} - 2 x + 1 + y^{2} - 4 y + 4 = 4 + 1 + 4$ $x^2-2x+1+y^2-4y+4=4+1+4$

Factorise

${(x - 1)}^{2} + {(y - 2)}^{2} = 9 = 3^{2}$ $(x-1)^2+(y-2)^2=9=3^2$

This is the equation of a circle, center $C = (1, 2)$ $C=(1,2)$ and radius $r = 3$ $r=3$

The graph is shown below

graph{(x^2+y^2-2x-4y-4)=0 [-8.065, 7.735, -2.18, 5.72]}

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