How do you graph $x^{2} + y^{2} + 6 x - 4 y - 12 = 0$ $x^2 + y^2 + 6x - 4y - 12 = 0$ ?

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How do you graph $x^{2} + y^{2} + 6 x - 4 y - 12 = 0$ $x^2 + y^2 + 6x - 4y - 12 = 0$ ?

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Answer 1 · 2016-03-01T16:03:20

circle : centre (-3,2) , r = 5

Explanation:

The general form of the equation of a circle is :

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ $x^2 + y^2 + 2gx + 2fy + c = 0$

centre = (-g,-f) and r $= \sqrt{g^{2} + f^{2} - c}$ $=sqrt(g^2 + f^2 -c)$

the equation $x^{2} + y^{2} + 6 x - 4 y - 12 = 0 is in this form$ $x^2 + y^2 + 6x - 4y - 12 = 0 " is in this form "$

and by comparison: 2g=6 →g=3, 2f=-4 →f=-2 and c = -12

centre = (-3,2) and r= $\sqrt{3^{2} + {(- 2)}^{2} + 12} = \sqrt{25} = 5$ $sqrt(3^2+(-2)^2+12) = sqrt25 = 5$

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How do you graph $x^{2} + y^{2} + 6 x - 4 y - 12 = 0$ $x^2 + y^2 + 6x - 4y - 12 = 0$ ?

1 Answer

Explanation:

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How do you graph $x^{2} + y^{2} + 6 x - 4 y - 12 = 0$ $x^2 + y^2 + 6x - 4y - 12 = 0$ ?