The graph is of which kind?

From the given equation:

$\begin{aligned} x^2 - 2x + 8y + y^2 & = 8 \\ x^2 - 2x + 1 + y^2 + 8y + 16 & = 8 + 1 + 16 \\ (x-1)^2 + (y+4)^2 & = 5^2 \end{aligned}$

It is a circle with center at $(1,-4)$ and radius $5$ .

Any equation of the form $x^2 + y^2 = r^2$ is a circle.

An equation of the form $(x-h)^2 + (y-k)^2 = r^2$ is also a circle, but centered at the point $(h,k)$ .

The equation above expands to:

$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2$

And simplifies to:

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

If we want the equation above to match the given equation, then the following must be true:

$-2hx = -2x$ $\rightarrow h=1$

$-2ky = 8y$ $\rightarrow k=-4$

Plugging our h and k values into the expanded equation gives:

$x^2 - 2x +8y + y^2 = r^2 - (17)$

This matches the given equation if $r^2-17=8$ , in which case $r^2 = 25$ .

So, the given equation is a circle of radius $5$ centered at $(1,-4)$ .