Prove that midpoints of chords passing through a certain point lie on a circle

Question

Let $P$ be a point inside circle $C_1$.Consider the set of chords of $C_1$ that contain $P$. Prove that their midpoints all lie on a circle.

My observation: Let $O$ be the centre of $C_1$.$O$ must in the required circle. To show some points concyclic usually we need angles.I tried to find angled at the midpoints but failed to find any angle. Please help me. Thank you!

Guess It is just my guess but constructing a good diagram I found that the circle may be the circle with diameter $OP$.

Answer 1

Let $O$ be the centre of $C_1$, Let $A$ be the midpoint of a cord passing through $P$. Since $A$ is the midpoint of the chord, $\angle OAP =90^{\circ}$. Thus the circumcircle of $\Delta OAP$ has diameter $OP$, which is fixed for all chords. Thus all the midpoints lie on a circle with diameter $OP$

Answer 2

Set a $P$ and a circle in coordinate system. Say $P(0,2n)$ and $$C_1:\;\; x^2+y^2=1$$ Say arbitrary line through $P$: $\;\;y =kx+2n$ cuts in $A(x_1,y_1)$ and $B(x_2,y_2)$. Let $M(x',y')$ be a midpoint of $AB$. Then we get $x_i$ from $$(kx+2n)^2+x^2=1\;\;\; \Longrightarrow \;\;\; (k^2+1)x^2+4knx+4n^2-1=0$$

By Viete formulae we have $$x' ={1\over 2}(x_1+x_2) = -{2kn\over k^2+1}$$ and $$y' = kx'+2n $$

Finally we get $$x'^2+(y'-n)^2 =\underbrace{(k^2+1)x'^2+2knx'}_{=0}+n^2 =n^2 $$ So $M$ is on circle with center at $O(0,n)$ and radius $n$.