Htykuut

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?

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 $left(2h, 2k-5right)$.

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 $left(0,frac{5}{2}right)$

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 $C$.