How do you graph $x^{2} + y^{2} + 6 x - 2 y - 15 = 0$ $x^2+y^2+6x-2y-15=0$ ?

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How do you graph $x^{2} + y^{2} + 6 x - 2 y - 15 = 0$ $x^2+y^2+6x-2y-15=0$ ?

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Answer 1 · 2016-06-03T06:44:45

This is the equation of a circle with center $(- 3, 1)$ $(-3,1)$ and radius $5$ $5$

Explanation:

As in the equation $x^{2} + y^{2} + 6 x - 2 y - 15 = 0$ $x^2+y^2+6x-2y-15=0$ ,

coefficients of $x^{2}$ $x^2$ and $y^{2}$ $y^2$ are equal and there is no term having $x y$ $xy$ (in other words coefficient of $x y$ $xy$ is $0$ $0$ ),

hence this is the equation of a circle.

Now $x^{2} + y^{2} + 6 x - 2 y - 15 = 0$ $x^2+y^2+6x-2y-15=0$

$\Leftrightarrow (x^{2} + 6 x + 9) + (y^{2} - 2 y + 1) = 15 + 9 + 1$ $hArr(x^2+6x+9)+(y^2-2y+1)=15+9+1$

$\Leftrightarrow {(x + 3)}^{2} + {(y - 1)}^{2} = 5 \cdot 2$ $hArr(x+3)^2+(y-1)^2=5*2$

It is therefore a circle with center $(- 3, 1)$ $(-3,1)$ and radius $5$ $5$

graph{x^2+y^2+6x-2y-15=0 [-12.42, 7.58, -3.84, 6.16]}

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How do you graph $x^{2} + y^{2} + 6 x - 2 y - 15 = 0$ $x^2+y^2+6x-2y-15=0$ ?

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How do you graph $x^{2} + y^{2} + 6 x - 2 y - 15 = 0$ $x^2+y^2+6x-2y-15=0$ ?