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

As a part of an exercise It would help me if I could prove formally that the set $\{ (x,y) \in R^2 \mid x^2 + y^2 -2x + 4y - 11 = 0 \}$ is closed and bounded.

Plotting it with a software I can see this immediately but I am bit doubtful on a formal proof, any help?

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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 $\{(x,y) \mid \|(x,y)\| \le R \}$.

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Thank you a lot for your answer, I got why it's closed. But why would $\lim_{\|(x,y)\| \to \infty} f(x,y) = \infty$ imply it's bounded? – Monolite Jan 18 at 19:33
I think he's trying to say that if it were unbounded, i.e. $(x,y)$ were unbounded, then $f(x,y)$ would also be unbounded, which certainly means it can't be zero. Basically, your set is a circle of finite radius, which will be bounded. – Future Jan 18 at 19:38
@Monolite: I added a clarification. – copper.hat Jan 18 at 19:38

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

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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?

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So I choose a point $(x,y)$ then I look at the distance from zero $d((x,y),0)$ so I have $\sqrt(x^2 + y^2) < ?$, hum no. for the boundness I am having some problems, but I can prove that the complement is open. Thanks. – Monolite Jan 18 at 19:44
Could you help me with the length argument? I am trying to think about it but it's not coming to me. – Monolite Jan 18 at 20:05
From there I can say $\sqrt{11 + 2x -4y} \le \sqrt{45} = 3 \sqrt{5}$ ? I am a bit unsure. – Monolite Jan 18 at 20:41
But this is an arbitrary point that does not necessarily belong to my set. – Monolite Jan 18 at 20:42
You have to use newly the hypothesis that $(x,y)$ belongs to your set – sinbadh Jan 18 at 20:48