$\{ (x,y) \in R^2 \mid x^2 + y^2 -2x + 4y - 11 = 0 \}$ is closed and bounded

Since the function $f(x,y) = x^2 + y^2 -2x + 4y - 11$ is continuous, it follows that $f^{-1}(\{0\})$ is closed.

Since $ \lim_{\|(x,y)\| \to \infty} f(x,y) = \infty$, we see that the set must be bounded.

To see why this implies that the set is bounded, note that we can find some $R$ such that if $\|(x,y)\| > R$, then $f(x,y) > 1$. Hence the set $\{(x,y) \mid f(x,y) \le 1 \}$ is contained in $\{(x,y) \mid \|(x,y)\| \le R \}$.

Hint $(x-1)^2+(y+2)^2=16$ is the set so it is a circle and bounded and closed

As you seen, it is a circle with certain center $x_0=(1,-2)$ and radio $r=4$.

For the boundness, show that, every point in your set has lenght $<7\sqrt{2}$.

For the closeness, show that the complement is open (you only have to follow the geometry of your set).

Can you continue?

Edit

$\begin{eqnarray} \sqrt{x^2+y^2}&=&\sqrt{(x-1+1)^2+(y+2-2)^2}\\ &=&\sqrt{(x-1)^2+2(x-1)+1+(y+2)^2-4(y+2)+4}\\ &=&\sqrt{16+2(x-1)-4(y+2)+5}\\ &=&\sqrt{21+2x-2-4y-8}\\ &=&\sqrt{11+2x-4y} \end{eqnarray}$

Newly, can you continue?