Lösung 4.1:7d
Aus Online Mathematik Brückenkurs 1
We rewrite the equation in standard by completing the square for the x- and y-terms:
\displaystyle x^{2}-2x=\left( x-1 \right)^{2}-1^{2}
\displaystyle y^{2}+2y=\left( y+1 \right)^{2}-1^{2}
Now, the equation is
\displaystyle \begin{align}
& \left( x-1 \right)^{2}-1+\left( y+1 \right)^{2}-1=-2 \\
& \Leftrightarrow \quad \left( x-1 \right)^{2}+\left( y+1 \right)^{2}=0 \\
\end{align}
The only point which satisfies this equation is
\displaystyle \left( x \right.,\left. y \right)=\left( 1 \right.,\left. -1 \right)
because, for all other values of
\displaystyle x
and
\displaystyle y , the left-hand side is strictly positive and therefore not zero.
