How do you graph $x^{2} + y^{2} + 8 x - 6 y + 16 = 0$ $x^2 + y^2 + 8x - 6y + 16 = 0$ ?

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How do you graph $x^{2} + y^{2} + 8 x - 6 y + 16 = 0$ $x^2 + y^2 + 8x - 6y + 16 = 0$ ?

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Answer 1 · 2016-04-11T13:05:08

$x^{2} + y^{2} + 8 x - 6 y + 16 = 0$ $x^2+y^2+8x-6y+16=0$
is a circle with radius $3$ $3$ and center $(- 4, 3)$ $(-4,3)$

Explanation:

$x^{2} + y^{2} + 8 x - 6 y + 16 = 0$ $x^2+y^2+8x-6y+16=0$

$\Rightarrow (x^{2} + 8 x + 16) + (y^{x} - 6 y + 9 = 0 + 9$ $rArr (x^2+8x+16)+(y^x-6ycolor(red)(+9)=0color(red)(+9)$

$\Rightarrow {(x + 4)}^{2} + {(y - 3)}^{2} = 3 \land 2$ $rArr (x+4)^2+(y-3)^2=3^^2$
$XXX$ $color(white)("XXX")$ which is the general form for a circle with center $(- 4, 3)$ $(-4,3)$ and radius $3$ $3$
graph{x^2+y^2+8x-6y+16=0 [-9.636, 4.41, -0.826, 6.196]}

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How do you graph $x^{2} + y^{2} + 8 x - 6 y + 16 = 0$ $x^2 + y^2 + 8x - 6y + 16 = 0$ ?

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Explanation:

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How do you graph x^2 + y^2 + 8x - 6y + 16 = 0x2+y2+8x−6y+16=0x^2 + y^2 + 8x - 6y + 16 = 0?

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How do you graph $x^{2} + y^{2} + 8 x - 6 y + 16 = 0$ $x^2 + y^2 + 8x - 6y + 16 = 0$ ?