Triangle construction procedure

Question

Two lines $L1,L_2$ pass through a common point $O. $ $L_2$ goes through points $P$ and $Q$. How to construct a circle through $P,Q$ to be tangent to $L_1?$

In a particular case, at the tangent point $T$ we have similar triangles $OTP, OQT. $

Answer 1

The main point is to locate $X$ (which is your $T$) on $L_1$ such that it is the point of contact of the required circle that passes through $P$ and $Q$ on $L_2$.

By “power of a point”, $OX^2 = OP.OQ$. The RHS and the LHS of this equation are respectively a rectangle whose sides are $OP \times OQ$ and a square whose sides are $OX$ (and the 2 figures are equal in area).

The way of finding $OX$ is:-

1. By drawing the red dotted circle, locate $Q’$ on $PO$ extended such that $OQ = OQ’$.

2. On $OQ’$, locate $M$, the midpoint of $PQ’$.

3. Using $M$ as center with $MP$ as radius, draw the green dotted circle.

4. AT $O$, erect the perpendicular $OX’$ cutting the green dotted circle at $X’$.

After the above construction, the power of $O$ w.r.t. the green dotted circle is $OX’^2 = OQ’.OP$.

The final step is to draw the blue dotted circle (centered at $O$, $OX’$ as radius). It cuts $L_1$ at $X$.

$XPQ$ is the required circle.

The answer to your last question is yes (and always).

The blue dotted circle cuts $L_1$ at point $Y$ too. The black dotted circle passing through $Y, P, Q$ also meets the requirement. Below is an updated version according to @RoryDaulton 's suggestion:-