\section*{Problem 3 - V11133}

Given are two fixed points \(A\) and \(B\) with the distance \(e\).

a) Where do all points \(F\) lie for which the squares of their distances from \(A\) and \(B\) have the fixed sum \(s\)?

b) Are there such points for any choice of \(e\) and \(s\)?
}

a) O.B.d.A. let \(A=\left(-\frac{e}{2}, 0\right)\) and \(B=\left(\frac{e}{2}, 0\right)\), then all points \(F=(x, y)\) with

\[
s=\left(\left(x+\frac{e}{2}\right)^{2}+y^{2}\right)+\left(\left(x-\frac{e}{2}\right)^{2}+y^{2}\right)=2 x^{2}+2 \cdot \frac{e^{2}}{4}+2 y^{2}
\]

are sought. This equation is equivalent to the circle equation

\[
x^{2}+y^{2}=\frac{s}{2}-\frac{e^{2}}{4}
\]

The sought points \(F\) thus form a circle whose center is the midpoint of the segment \(A B\) and whose radius is \(\frac{s}{2}-\frac{e^{2}}{4}\).

b) Since squares of real numbers are always greater than or equal to zero, there are only such points \(F\) if \(\frac{s}{2}-\frac{e^{2}}{4} \geq 0\). In particular, there are not such points for every choice of \(e\) and \(s\).