The circle with equation x^2 + 4x + y^2 - 6y = 3 has center C and radius r, where?
- C = (0,0) ; r = Ö3
- C = (2, -3) ; r = 16
- C = (2, -3) ; r = 4
- C = (-2,3) ; r = 16
- C = (-2,3) ; r = 4
The circle's equation is:
$\displaystyle x^2 + 4x + y^2 - 6y = 3$
Rearrange this using the "identities"
$\displaystyle (x+2)^2 \equiv x^2 + 4x + 4$
$\displaystyle (y-3)^2 \equiv y^2 - 6y + 9$
and you'll find it changes to:
$\displaystyle (x+2)^2 - 4 + (y-3)^2 - 9 = 3$
$\displaystyle \Rightarrow (x+2)^2 + (y-3)^2 = 16$
Compare it with the general form for a circle, radius $\displaystyle R$ centred at $\displaystyle (a,b)$:
$\displaystyle (x-a)^2 + (x-b)^2 = R^2$
and you can see that $\displaystyle R =4$, and the centre is $\displaystyle (-2, 3)$.