Proving System of Equations can be solved about $(3,0,1)$

Question

$x^2+y^2+z^2-6\sqrt{x^2+y^2}+8=0$ $(i)$

$x^2+y^2+z^2-6x-2y+8=0$ $(ii)$

Prove: The system of equations can be solved in a neighborhood of $(3,0,1)$ through unique continuous functions $x\to y(x)$ and $x \to z(x)$.

This is the first time I've seen a question like this, so I am clueless, but my ideas:

putting $(i)$ in $(ii)$, we get:

$6\sqrt{x^2+y^2}-8-6x-2y+8=0 \Rightarrow 36(x^2+y^2)=6x+2y$ and then somehow I need to get write $x$ as a function of $y$, but then how can I show that it is continuous and particularly unique? What role does the neighborhood $(3,0,1)$ play in all of this?

Any help is greatly appreciated.

Answer 1

Equating the $2$ equations gives

$$9(x^2+y^2)=(3x+y)^2$$

$$9x^2+9y^2=9x^2+6xy+y^2$$

$$8y^2=6xy$$

$$y(y-\frac{3}{4}x)=0$$

Hence either $y=0$ of $y=\frac{3}{4}x$

Note that we are allowed to square the equation in a small neighbourhood of $(3,0,1)$ as $6x+2y>0$ there.

Now consider the point $(3,0,1)$ - clearly $y=0$. But as $x \neq0$ this implies that in a small neighbourhood of that point $y=0$ by the continuity of the solution - it can't be $y=\frac{3}{4}x$ since at $x=3$ we would have $y=\frac{9}{4}$.

Then in this small neighbourhood we have $y=0.$

Plugging this in gives that in a small neighbourhood around the point we have $$x^2-6x+8+z^2=0$$ $$(x-3)^2+z^2=1$$

which is a circle centred at $3$ with radius $1$ in the $xz$-plane.

Then use continuity again.

Answer 2

This problem is a rather straightforward application of the Implicit function theorem.

I recommend you read the article (or check your notes) and then ask again if you need more hints.

If you need a little more help, try inserting (3, 0, 1) into the equation and think about how that might relate to the linked theorem.

Answer 3

When you identify the equation you have that

$6\sqrt{x^2+y^2}=6x+2y$

$\sqrt{x^2+y^2}=x+\frac{1}{3}y$

So for $(x,y)\in \mathbb{R}^2$ such that $6x+2y\geq0$ you have that they are solutions of your equation if and only if

$ x^2+y^2=x^2+\frac{1}{9}y^2+\frac{2}{3}xy$

Then

$\frac{2}{3}xy=\frac{8}{9}y^2$

so

$xy=\frac{4}{3} y^2$

If $y\neq 0$ than $x=\frac{4}{3}y$

In any case this not help you to find an answer for your question, the idea is to apply the implicit function theorem.