the circle equation

**Alexander** · June 6th 2007, 04:03 AM

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

**Pterid** · June 6th 2007, 04:43 AM

The circle's equation is:

$x^2 + 4x + y^2 - 6y = 3$

Rearrange this using the "identities"
$(x+2)^2 \equiv x^2 + 4x + 4$
$(y-3)^2 \equiv y^2 - 6y + 9$

and you'll find it changes to:

$(x+2)^2 - 4 + (y-3)^2 - 9 = 3$
$\Rightarrow (x+2)^2 + (y-3)^2 = 16$

Compare it with the general form for a circle, radius $R$ centred at $(a,b)$ :

$(x-a)^2 + (x-b)^2 = R^2$

and you can see that $R =4$ , and the centre is $(-2, 3)$ .

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