Proof for a nice conjecture related to circles I have made a conjecture but am unable to prove it.

If $P$ and $Q$ are any points such that they are both at an equal given distance from a given point $C$, subtending a fixed angle at another given point $O$ (wherein $ OC<PC=QC $). Then the maximum and minimum distance $PQ$ occurs when $P$ and $Q$ lie on equally inclined rays to the line perpendicular to $OC$ through $O$.

Clearly, the minimum case would be in the smaller segment and maximum in the larger segment. Also note  that if $P$ and $Q$ subtend an angle $2x$ at $O$, then each of the rays will be inclined at $90°-x$ to the line perpendicular to $OC$.
 A: For convenience, first rename $C$ and $O$ respectively as $O$ and $R$, and then choose the coordinate frame so that $O$ is the origin and $R$ is the point $(c,0)$, where $0<c<1$, while $P$ is at $(\cos(\theta+\phi),\sin(\theta+\phi))$ and $Q$ is at $(\cos(\theta-\phi),\sin(\theta-\phi))$, so that $|OP|=|OQ|=1$. The tangent of the angle $\angle PRQ$, which is constant, can be obtained from the gradients $\tan\alpha$ and $\tan\beta$ of $RP$ and $RQ$ respectively, according to the formula
$$\tan\angle PRQ=\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}.$$
Here we substitute
$$\tan\alpha=\frac{\sin(\theta+\phi)}{\cos(\theta+\phi)-c}\quad\text{and}\quad\tan\beta=\frac{\sin(\theta-\phi)}{\cos(\theta-\phi)-c}$$
and simplify using standard trigonometric formulae to get
$$\frac{\sin2\phi-2c\cos\theta\sin\phi}{\cos2\phi-2c\cos\theta\cos\phi+c^2}=\tan\angle PRQ\text{ (= constant).}$$
Now take the logarithm and differentiate (indicated by a prime) with respect to $\theta$, yielding
$$\frac{2\phi'\cos2\phi+2c(\sin\theta\sin\phi-\phi'\cos\theta\cos\phi)}{\sin2\phi-2c\cos\theta\sin\phi}$$
$$=\frac{2\phi'\sin2\phi+2c(\sin\theta\cos\phi+\phi'\cos\theta\sin\phi)}{\cos2\phi-2c\cos\theta\cos\phi+c^2}.$$
Now $|PQ|=2\sin\phi$, and for this to be minimized, with $0\leqslant\phi\leqslant\frac12\pi$, we must have
$\phi'=0$, so that
$$\frac{\sin\theta\sin\phi}{\sin2\phi-2c\cos\theta\sin\phi}=\frac{\sin\theta\cos\phi}{\cos2\phi-2c\cos\theta\cos\phi+c^2}.$$ Then either $\theta=0$ (giving the required result) or, after division by $\sin\theta$, we have $c^2=1$, which is not permissible.
A: Consider the circle $a$ through $OPQ$ (violet in figure below). If this circle is not tangent to circle $c$ of center $C$ passing through $O$ (dashed), then on $c$ there are points both inside $a$ (as $O'$ in the figure) and outside $a$ (as $O''$). But then
$$
\angle PO''Q<\angle POQ<\angle PO'Q
$$
and from that it follows that it's possible to construct an angle of vertex $O'$ equal to $\angle POQ$ but subtending an arc on $a$ smaller than $PQ$, and it's possible to construct an angle of vertex $O''$ equal to $\angle POQ$ but subtending an arc on $a$ larger than $PQ$.
But the same angles can be constructed from vertex $O$ (just rotate them about $C$), hence if $a$ is not tangent to $c$ then point $O$ cannot subtend a minimum or maximum arc.
It follows that point $O$ subtends a minimum or maximum arc only if circle $POQ$ is tangent to circle $c$, which is equivalent to the thesis to be proved.

