How do you graph $x^{2} + y^{2} - 6 x + 2 y - 6 = 0$ $x^2 + y^2 -6x + 2y - 6 = 0$ ?

Question

14399 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-26T11:50:00

Draw a circle with center $(3, - 1)$ $(3,-1)$ and radius $4$ $4$ .

$x^{2} + y^{2} - 6 x + 2 y - 6 = 0$ $x^2+y^2-6x+2y-6=0$

$\to x^{2} - 6 x + y^{2} + 2 y = 6$ $rarr color(red)(x^2-6x)+color(blue)(y^2+2y)=color(green)(6)$

$\to x^{2} - 6 x + 9 + y^{2} + 2 y + 1 = 6 + 9 + 1$ $rarr color(red)(x^2-6x+9)+color(blue)(y^2+2y+1) = color(green)(6)color(red)(+9)color(blue)(+1)$

$\to {(x - 3)}^{2} + {(y + 1)}^{2} = 4^{2}$ $rarr color(red)((x-3)^2) + color(blue)((y+1)^2) = color(orange)(4^2)$

which is the standard form for a circle with center $(3, - 1)$ $(3,-1)$ and radius $4$ $4$ .

graph{x^2-6x+y^2+2y-6=0 [-8.1, 9.68, -5.404, 3.48]}

How do you graph x^2 + y^2 -6x + 2y - 6 = 0x2+y2−6x+2y−6=0x^2 + y^2 -6x + 2y - 6 = 0?