# Prove that the common chord passes through the origin.

Consider the parabola $x^2=4ay$. Two focal chords are constructed, and the interval that the chord makes with the parabola are diameters of two circles.

Prove that that the common chord of the two circles passes through the origin (vertex of the parabola).

I am having trouble finding a purely geometric way of proving this result.

• That's a neat theorem. – marty cohen Jan 18 '18 at 1:19
• I have a coordinate proof of the following: Given a circle whose diameter is a focal chord of the parabola, the line tangent to the parabola at its vertex ($V$) meets the circle at $P$ & $Q$ such that $|\overline{PV}| |\overline{QV}| = -3a^2$ (using "signed" distances). That is, the power of $V$ with respect to the circle is $-3a^2$, a value that depends only upon the parabola itself. Thus, given any two such circles, $V$ has the same power with respect to each, so that it must lie on their radical axis, ie, their extended common chord. – Blue Jan 18 '18 at 2:16

Consider a parabola with focus $F$ and vertex $V$; define $a := |\overline{VF}|$. Let $\overline{PQ}$ be a focal chord of the parabola, with $M$ its midpoint. Let $F^\prime$, $P^\prime$, $Q^\prime$, $M^\prime$ be the projections of the corresponding points onto the directrix. (Note that $|\overline{VF^\prime}| = a$.)
It is "known" that the tangents (not shown) at the endpoints of a focal chord are perpendicular, and that they meet at the point on the directrix halfway between their own projections. In our scenario, that point must be $M^\prime$, so that the circle with diameter $\overline{PQ}$ is tangent to the directrix at $M$. Let $r := |\overline{MM^\prime}|$ be the radius of that circle.
It is also "known" that the tangents at $P$ and $Q$ bisect respective angles $\angle FPP^\prime$ and $\angle FQQ^\prime$. This implies that $F$ is the common reflection of $P$ over $\overline{PM^\prime}$ and $Q$ over $\overline{QM^\prime}$, so that $\overline{FM^\prime}\perp\overline{PQ}$. Define $m := |\overline{FM^\prime}|$. A little angle chasing shows that $\angle FM^\prime F^\prime \cong \angle FMM^\prime$, so that the similar right triangles yield $$\frac{2a}{m} = \frac{m}{r} \qquad\to\qquad m^2 = 2 a r \tag{1}$$
This comes in handy for calculating the power of $V$ with respect to the circle:
\begin{align} \text{power of V wrt \bigcirc{M}} &:= n^2 - r^2 \\ &\,= n^2 - (r - a)^2 - 2 a r+ a^2 \\ &\,= |\overline{M^\prime F^\prime}|^2 - 2 a r + a^2 \\ &\,= m^2 - (2a)^2 - 2 a r + a^2 \\ &\,= (m^2 - 2 a r) - 3 a^2 \\ &\,= - 3 a^2 \tag{2} \end{align}
We observe that this value is independent of our choice of $P$ and $Q$, and is therefore a constant of this configuration. Consequently, for any two focal-chord-diameter circles, vertex $V$ has the same power with respect to each; this places $V$ on the circles' radical axis, which for intersecting circles is the line containing their common chord. $\square$