$\displaystyle x^2+y^2-2x-4y+6=0$ 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 $\displaystyle x^2+y^2+2gx+2fy+c=0$
Thanks
$\displaystyle x^2+y^2-2x-4y+6=0$ 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 $\displaystyle x^2+y^2+2gx+2fy+c=0$
Thanks
$\displaystyle x^2 - 2x + y^2 - 4y = -6$
complete the square for x and y ...
$\displaystyle x^2 - 2x + 1 + y^2 - 4y + 4 = -6 + 1 + 4$
$\displaystyle (x-1)^2 + (y-2)^2 = -1$
since the general circle equation is
$\displaystyle (x-h)^2 + (y-k)^2 = r^2$
$\displaystyle r^2 = -1$ ?
Actually, the "general equation" of a circle is of the form
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$
which is a circle of radius $\displaystyle r$, centred at $\displaystyle (h, k)$.
I would try to transform the equation you have been given into this form.
$\displaystyle x^2 + y^2 - 2x - 4y + 6 = 0$
$\displaystyle x^2 - 2x + y^2 - 4y = -6$
$\displaystyle x^2 - 2x + (-1)^2 + y^2 - 4y + (-2)^2 = -6 + (-1)^2 + (-2)^2$
$\displaystyle (x - 1)^2 + (y - 2)^2 = -1$.
Is $\displaystyle r^2 = -1$ possible?