Is this problem a general for of the equation of a circle?

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Nov 12th 2008, 06:59 PM
scottydoint

Is this problem a general for of the equation of a circle?

-x^2-4x+4y^2+8y-22=0
Nov 12th 2008, 07:27 PM
Soroban

Hello, scottydoint!

Quote:

$-x^2-4x+4y^2+8y-22\:=\:0$

It is not a circle . . . $x^2$ and $y^2$ have different coefficient.

Since the coefficients of $x^2$ and $y^2$ have opposite signs, it is a hyperbola.
Nov 12th 2008, 07:38 PM
earboth

Quote:

Originally Posted by scottydoint

-x^2-4x+4y^2+8y-22=0

1. Complete the squares:

$-(x^2+4x{\bold{\color{red}+4}})+4(y^2+2y{\bold{\col or{blue}+1}})=22 {\bold{\color{red}-4}} {\bold{\color{blue}+4}}$

2. Re-write:

$-x^2-4x+4y^2+8y-22=0~\implies~ -\dfrac{(x+2)^2}{22} + \dfrac{(y+1)^2}{5.5} =1$

3. This is the equation of a hyperbola opening up-down.

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