## Task 1 - 311221

If $c$ is a real number, the system of equations

$$
\begin{aligned}
x+y & =1 \\
x^{2}+y^{2} & =c
\end{aligned}
$$

is formed.

a) Determine all pairs $(x ; y)$ that satisfy the system of equations (1), (2) for $c=2$.

Furthermore, determine all real numbers $c$ for which the system of equations (1), (2)

b) has no solution $(x ; y)$ from real numbers $x, y$,

c) has exactly one solution $(x ; y)$,

d) has two different solutions $\left(x_{1} ; y_{1}\right),\left(x_{2} ; y_{2}\right)$ from real numbers.
Solving (1) for $y$ and substituting into (2) yields the quadratic equation

$$
x^{2}+(1-x)^{2}=2 x^{2}-2 x+1=c \Rightarrow x_{1,2}=\frac{1 \pm \sqrt{2 c-1}}{2}
$$

with the two pairs $(x, y)$ :

$$
\left(\frac{1-\sqrt{2 c-1}}{2} ; \frac{1+\sqrt{2 c-1}}{2}\right) \quad ; \quad\left(\frac{1+\sqrt{2 c-1}}{2} ; \frac{1-\sqrt{2 c-1}}{2}\right)
$$

For $c=2$, this results in $(x ; y) \in\left\{\left(\frac{1}{2}+\frac{\sqrt{3}}{2} ; \frac{1}{2}-\frac{\sqrt{3}}{2}\right) ;\left(\frac{1}{2}-\frac{\sqrt{3}}{2} ; \frac{1}{2}+\frac{\sqrt{3}}{2}\right)\right\}$.

The number of different solutions is determined by the radicand of $\sqrt{2 c-1}$.

No solution exists for $2 c-1<0$ (which simplifies to $c<\frac{1}{2}$).