How do you graph $x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ $x^2+y^2+6x-8y+9=0$ ?

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How do you graph $x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ $x^2+y^2+6x-8y+9=0$ ?

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Answer 1 · 2016-02-12T20:50:16

circle: centre(-3 , 4) and radius = 4

Explanation:

the general equation of a circle is

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

the equation here : $x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ $x^2 + y^2 + 6x - 8y + 9 = 0$
compares with the general equation and is therefore a circle.

by comparison of the 2 equations :

obtain: 2g = 6 → g = 3 , 2f = -8 → f = -4 and c = 9

centre = (-g , -f ) = (-3 , 4 )

and radius = $\sqrt{g^{2} + f^{2} - c} = \sqrt{{(- 3)}^{2} + 4^{2} - 9} = \sqrt{16} = 4$ $sqrt(g^2 + f^2 - c) = sqrt((-3)^2+4^2-9) = sqrt16 = 4$

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How do you graph $x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ $x^2+y^2+6x-8y+9=0$ ?

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How do you graph $x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ $x^2+y^2+6x-8y+9=0$ ?