analytic geometryeuclidean-geometrygeometry
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 $OP$.
The center of the circle is at the intersection of the perpendicular bisectors of the two chords. Any of the segments joining this point to an endpoint of a chord will be a radius.
Notice that transformation $A\mapsto B$ from bigger circle which has radius $2\sqrt{10}$: $$XY = \sqrt{4^2 +12^2} = 4\sqrt{10}$$ and center at $O(0,1)$ to smaller circle with radius $\sqrt{10}$ and center at $O'(3,-8)$ is homothety with center at $Y$ and dilatation factor $-1/2$. So, for all $A$ we have $${BY\over YA} = {1\over 2}\implies {AB\over YA} ={3\over 2}$$
Then $${YM \over YA} = 1- {AM\over YA} = 1-{AB\over 2YA} = {1\over 4} $$
So $M$ is actually a picture of $A$ under homothety at $Y$ with dilatation factor $1/4$, so $M$ descibes a circle with radius $r=\sqrt{10}/2$ and center at $({1\over 2},-{7\over 2})$.
Best Answer
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 $OP$