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Two circles intersect at $A$ and $B$; $PQ$ is a straight line through $A$ meeting the circles at $P$ and $Q$. Find the locus of midpoint of $PQ$.

I know the locus is a circle. But I am unable to prove it.

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Angles $\angle BPA$ and $\angle BQA$ don't depend on the positions of $P$ and $Q$, while they vary on the outer parts of the circles. That means that all triangles $BQP$ have the same angles, independent of the positions of $P$ and $Q$, and that is true even if $P$ and $Q$ lie on the inner parts of the circles.

If $M$ is the midpoint of $PQ$, it follows that median $BM$ forms a constant angle $\alpha$ with $PQ$. We then have $\angle AMB=\alpha$ if $PM<PA$, and $\angle AMB=\pi-\alpha$ if $PM>PA$: point $M$ lies then on the circle having $AB$ as a chord, subtended by an angle $\alpha$.

In particular, when $PQ$ is perpendicular to $AB$, the centers of the given circles are the midpoints of $BP$ and $BQ$, while the center of the locus is the midpoint of $BM$. It follows that the center of the locus is the midpoint of the centers of the given circles.

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  • $\begingroup$ Very useful. Thanks $\endgroup$ – Dayal Kumar May 15 '17 at 14:48
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Depends how you want to prove this, but going to polar coordinates with origin $A$ makes for a pretty straightforward proof.

"Polar coordinates"
For any point $x\in\mathbb R^2$ there exist $\rho\in\mathbb R$ and a unit vector $\mathbf e$ such that $x=A+\rho\mathbf e$. Note that I don't restrict $\rho$ to be positive so a point can be expressed by several different values of $(\rho,\mathbf e)$. It doesn't really matter since we can still express all points under that form.

Circle equation
Now consider some circle with center $c$ and radius $r$, such that $A$ belongs to the circle. The circle is the collection of points $x=A+\rho\mathbf e$ such that $$ \rho=r\left\langle\mathbf e,\ \frac{c-A}{\left\|c-A\right\|}\right\rangle =\left\langle\mathbf e,\ c-A\right\rangle $$

Locus
Now let's say you have two circles with centers $c_1$ and $c_2$, and radii $r_1$ and $r_2$. One line $PQ$ corresponds to a fixed direction $\mathbf e$. I assume that whenever possible, $P\neq A\neq Q$. This means that $P$ and $Q$ are the points $x_1=A+\rho_1\mathbf e$ and $x_2=A+\rho_2\mathbf e$, where $\rho_i=\langle\mathbf e,\ c_i-A\rangle$. The midpoint of $P$ and $Q$ is $A+\rho\mathbf e$ where $\rho=\frac{\rho_1+\rho_2}2$. It's easy to see there is $c$ such that $\rho=\langle\mathbf e,\ c-A\rangle$, that suffices to prove your locus is the circle of center $c$ and radius $r=\|c-A\|$.

In detail: \begin{align*} \rho &= \left\langle \mathbf e,\ \frac{c_1-A+c_2-A}{2} \right\rangle =\left\langle \mathbf e,\ \frac{c_1+c_2}{2}-A \right\rangle =\left\langle \mathbf e,\ c-A \right\rangle \end{align*} So just take $c$ as the midpoint between $c_1$ and $c_2$, your locus is the circle centered at $c$ that goes through $A$.

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