True or false?

**arze** · October 5th 2009, 05:03 PM

$x^2+y^2-2x-4y+6=0$ is the equation of a circle.

I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle $x^2+y^2+2gx+2fy+c=0$
Thanks

**skeeter** · October 5th 2009, 05:43 PM

Originally Posted by arze

$x^2+y^2-2x-4y+6=0$ is the equation of a circle.

I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle $x^2+y^2+2gx+2fy+c=0$
Thanks

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

complete the square for x and y ...

$x^2 - 2x + 1 + y^2 - 4y + 4 = -6 + 1 + 4$

$(x-1)^2 + (y-2)^2 = -1$

since the general circle equation is

$(x-h)^2 + (y-k)^2 = r^2$

$r^2 = -1$ ?

**Prove It** · October 5th 2009, 05:44 PM

Originally Posted by arze

$x^2+y^2-2x-4y+6=0$ is the equation of a circle.

I answered true, but the answer in the book is false. So who is correct? This equation satisfies the general equation of a circle $x^2+y^2+2gx+2fy+c=0$
Thanks

Actually, the "general equation" of a circle is of the form

$(x - h)^2 + (y - k)^2 = r^2$

which is a circle of radius $r$ , centred at $(h, k)$ .

I would try to transform the equation you have been given into this form.

$x^2 + y^2 - 2x - 4y + 6 = 0$

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

$x^2 - 2x + (-1)^2 + y^2 - 4y + (-2)^2 = -6 + (-1)^2 + (-2)^2$

$(x - 1)^2 + (y - 2)^2 = -1$ .

Is $r^2 = -1$ possible?

**arze** · October 5th 2009, 05:46 PM

Ah!!! How could I have missed that. Thanks!

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