This question is a Conics/Locus problem:
The circle $x^2+y^2=25$ cuts the y axis above the x axis at A. Find the locus of the midpoints of all chords of this circle that have A as one endpoint.
I’ve reasoned that the answer will be a circle with radius 2.5, but I don’t know how to mathematically prove it. I assume you use the midpoint formula?
On
Construct green triangle $ OA = OB =R $ , $C$ is given as mid point of chord $ AB,$ so, the triangle $ OCA $ is right angled, being congruent to $OCB$. Now $R=5,$ Using polar coordinates and then Cartesian,
$$ \dfrac{r}{\sin \theta } = R = \dfrac{r^2}{r \sin \theta } = \dfrac{x^2+y^2}{y} $$
$$ x^2+y^2 -y \,R =0 $$
which is the equation of red circle as locus of $C$.
The given circle $x^2+y^2=25$ has radius $5$ and center $(0,0)$. So based on your description $A$ has coordinates $(0,5)$.
Let $M(h,k)$ be the mid-point of the chord whose one end is at point $A$ and the other end at point $P$ on the circle. Then coordinates of $P$ will be given by $\left(2h, 2k-5\right)$.
Since $P$ lies on the circle, so it should satisfy the equation of the given circle. This means \begin{align*} (2h)^2+(2k-5)^2&=25\\ 4h^2+4k^2-20k & =0\\ h^2+k^2-5k&=0 \end{align*} So the locus of these midpoints is the circle given by $x^2+y^2-5y=0$. This is a circle with radius $\frac{5}{2}$ and center at $\left(0,\frac{5}{2}\right)$