Consider two straight lines $L_{1}$ and $L_{2}$ intersecting at a point P and inclined at a fixed acute angle $\theta$. If a rod $AB$ of fixed length $l$ slides along $L_{1},L_{2}$ such that $A$ always lies on $L_{1}$ and $B$ always lies on $L_{2}$,then
(A) the locus of the circumcenter of the triangle PAB is a circle
(B) area of the locus of circumcenter of the triangle PAB is $\frac{1}{8}l^2\theta\ cosec^2\theta$
(C) the locus of circumcenter of the triangle PAB is the major segment of a circle
(D)area of the locus of circumcenter of the triangle PAB is $\frac{1}{2}l^2\theta\ cosec^2\theta$
My Attempt
Taking the analytical approach(method of co-ordinates) I assumed P to be the origin $(0,0)$ and the line $L_{1}$ to be the x-axis.
Further I took point A as $(\alpha,0)$ and B as $(\beta,\beta\tan \theta)$.
If $R(h,k)$ be the circumcenter then circumcircle can be assumed as
$x^2+y^2-2hx-2ky=0$.
On putting coordinates of A and B in the circle and using the fact that $AB=l$
I eliminated $\alpha$ and $\beta$ and managed to obtain the equation(it was lot of rough work)
$h^2+k^2=\frac{l^2 \ cosec^2 \theta}{4}$
which indicates that the circumcenter $R(h,k)$ of triangle PAB lies on the circle
$x^2+y^2=\frac{l^2 \ cosec^2 \theta}{4}$
But I am not able to determine whether the circumcenter lies on the major or minor arc. What can we say about area.
BTW, the sine law says that the circumradius $r$ of $\triangle PAB$ is given by
$$\begin{align*} 2r &= \frac{AB}{\sin \angle APB}\\ &= \frac{l}{\sin\theta}\quad\text{ or }\quad \frac{l}{\sin(\pi-\theta)}\\ r&= \frac{l \csc \theta}{2} \end{align*}$$
So the circumcentre is always at $\frac{l \csc \theta}{2}$ away from $P$, the origin, and that distance is independent of the position of $AB$.
This matches your progress, though it doesn't confirm whether the locus is the full circle. My earlier comment was that the locus would miss a countable few points of the circle, when the circumcentre of $\triangle PAB$ is not defined as either $A$ or $B$ can overlap with $P$.