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

• November 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
• November 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.

• November 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.