Why a point $P$ that maximizes $\angle OPQ$ must be a point of tangency of a circle with chord $OQ$?

Question

I've seen the following problem with solution:

Can someone explain:

• Why a point $P$ that maximizes $\angle OPQ$ must be a point of tangency of a circle with chord $OQ$?

• How do we construct this circle? It seems he takes a circle with center $\left(\frac{1}{2}, k\right)$ but it's not clear to me where this chord $OQ$ must be constructed or where this circle must be constructed so that $OQ$ is a chord.

Answer 1

(1) Let $P_1$ be a point on the tangent line but other than P.

Claim: There is always another point called $P_2$ (in between P and $P_1$) such that $\angle OP_2Q \ge \angle OP_1Q$.

Let the circle passing through $O,Q, P_2$ cut $QP_1$ at X. Then, $\theta = \theta’ \ge \phi$.

Note that $QP_2$ is shorter than $QP_1$ and the shortest that $QP_2$ can have is QP.

(2) The figure shows both P and P’ are points of tangencies but P’ does not meet the requirement and therefore should be rejected.

Answer 2

1. You can prove the tangency by bashing easily which I assume is something you are more than capable of. But the geometric argument would b to argue by contradiction. Suppose $P$ on the line $l$ gives the maximum angle for $\angle OPQ.$ If $l$ is not tangent to the circumcircle of $\triangle OPQ$ at $P,$ then there is another intersection $P'$ and it is very easy to see that for any $P''$ on this segment $P'P$ gives a larger value for $\angle OP''Q,$ which is a contradiction.
2. I do not know what you are asking here. $OQ$ is fixed and the author is finding the unique solution for $P$ by simply coordinate-bashing. Do you mean how to construct this $P$ by compass and straightedge? Even if you did, you can still reverse-engineer this construction from the algebraic solution with ease.