Showing a system of equations having two solutions in $\mathbb{R}^2$ 
Consider the system of equations in $\mathbb{R}^2$ \begin{align*}
\xi^2 + \eta^2 &= 4\\
e^{\xi} + \eta &= 1
\end{align*}
  Show that the system has two solutions in $\mathbb{R}^2$ has two solutions in  $\mathbb{R}^2$, one with $\xi > 0$ and $\eta < 0$ and one with $\xi < 0$ and $\eta > 0$.

Attempted solution - The two equations are a circle of radius $2$ center at the origin and exponential opening down and to the left. If we rewrite equation $2$ as $\eta = 1 - e^{\xi}$ and plug in to equation $1$ we get $$\xi^2 - 2e^{\xi} + e^{2\xi} = 3$$
Using Wolfram we find two solution: $\xi = - 1.81626$ and $\xi = 1.00417$. Hence for $\xi = 1.00417$ we have $\eta = -1.7296$ and when $\xi = -1.81626$ we have $\eta = .83737$ satisfying the condition in the problem.
However, I am curious if there is a way of obtaining an interval to where the solution exists given the two conditions without using software of some kind. Any suggestions is greatly appreciated.
 A: Note that they only want you to show that there are two solutions, not solve for the solutions exactly. I would probably try to show that there are at least two solutions using the intermediate value theorem. Then that there cannot be more from comparing derivatives.
A: From the first equation, we can write 
$$ \xi = 2 \cos(t),\ \eta = 2 \sin(t), 0 \le t \le 2\pi$$
The second equation then becomes
$$ e^{2\cos(t)} + 2 \sin(t) = 1 $$
A plot of the left side, call it $f(t)$, looks like this:

Note that $f(\pi/2) = 3 > 1 > e^{-2} = f(\pi)$, so the Intermediate Value Theorem shows there is at least one solution with $\pi/2 < t < \pi$: this will have $\xi < 0$ and $\eta > 0$.  Similarly, $f(3\pi/2) = -1 < 1 < f(2\pi) =  e^2$, so there is at least one solution with $3\pi/2 < t < 2\pi$ which will have $\xi > 0$ and $\eta < 0$. 
Proving that there are exactly two solutions will be trickier.
A: Let $f(\xi ) =\xi ^2-2e^{\xi}+e^{2\xi}-3$.  Note that $$f(-2)=1-\frac{2}{e^2}+\frac{1}{e^4}>0 $$ and $$f(\ln(2))=\ln(2)^2-3<0$$ and $$f(\ln(3))=\ln(3)^2>0$$  By the Intermediate Value Theorem, you have at least two roots.  
