How do you graph $x^{2} + y^{2} + 4 x - 4 y - 1 = 0$ $x^2 + y^2 + 4x - 4y - 1 = 0$ ?

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How do you graph $x^{2} + y^{2} + 4 x - 4 y - 1 = 0$ $x^2 + y^2 + 4x - 4y - 1 = 0$ ?

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Answer 1 · 2016-02-15T20:26:56

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

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 : $x^{2} + y^{2} + 4 x - 4 y - 1 = 0 is in this form$ $x^2 + y^2 + 4x - 4y - 1 = "0 is in this form"$

by comparison: 2g = 4 → g = 2 , 2f = -4 → f=-2 and c = -1

centre = (-g , -f) = (-2 , 2)

and r $= \sqrt{g^{2} + f^{2} - c} = \sqrt{2^{2} + {(- 2)}^{2} + 1} = \sqrt{9} = 3$ $= sqrt(g^2 + f^2 - c) =sqrt(2^2+(-2)^2+1 )=sqrt9 = 3$

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How do you graph $x^{2} + y^{2} + 4 x - 4 y - 1 = 0$ $x^2 + y^2 + 4x - 4y - 1 = 0$ ?

1 Answer

Explanation:

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How do you graph $x^{2} + y^{2} + 4 x - 4 y - 1 = 0$ $x^2 + y^2 + 4x - 4y - 1 = 0$ ?