How do you graph $x^{2} + y^{2} + 3 x - 6 y + 9 = 0$ $x^2+y^2+3x-6y+9=0$ ?

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How do you graph $x^{2} + y^{2} + 3 x - 6 y + 9 = 0$ $x^2+y^2+3x-6y+9=0$ ?

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Answer 1 · 2016-03-05T11:12:12

circle : centre $(- \frac{3}{2}, 3), radius = \frac{3}{2}$ $( -3/2 , 3 ) , " radius " = 3/2$

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 = $\sqrt{g^{2} + f^{2} - c}$ $sqrt(g^2+f^2 - c )$

$x^{2} + y^{2} + 3 x - 6 y + 9 = 0 is in this form$ $x^2 + y^2 + 3x - 6y + 9 = 0 " is in this form "$

and by comparison : 2g = 3 → g $= \frac{3}{2}$ $=3/2$ , 2f = -6 → f = -3, c=9

centre = $(- \frac{3}{2}, 3) and r = \sqrt{{(\frac{3}{2})}^{2} + {(- 3)}^{2} - 9} = \frac{3}{2}$ $(-3/2 ,3 )" and " r = sqrt((3/2)^2+(-3)^2-9)=3/2$

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How do you graph $x^{2} + y^{2} + 3 x - 6 y + 9 = 0$ $x^2+y^2+3x-6y+9=0$ ?

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How do you graph x^2+y^2+3x-6y+9=0x2+y2+3x−6y+9=0x^2+y^2+3x-6y+9=0?

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How do you graph $x^{2} + y^{2} + 3 x - 6 y + 9 = 0$ $x^2+y^2+3x-6y+9=0$ ?