Two fixed points $A(p, q)$ and $B(r, s)$ and a fixed line $L: ax + by + c = 0$ are given. A variable point $P$ moves on the plane such that $AP$ intersects $L$ at $C$ and $BP$ intersects $L$ at $D$. What is the locus of $P$ if $CD$ is a fixed distance $d$?
My attempts at this are the following.
Take $P$ as $(h, k)$. Write the equation of the line in its parametric form. Name $C$ as $(f, g)$. Then in usual notation $D$ is $(f + d \cos\theta, g + d \sin\theta)$. Now using the formula for the area of a triangle (the determinant one), I set area of $APC$ as zero and area of $BPD$ as zero.
I cannot proceed further. Please help how to avoid the hairy algebra after this step or if there is any alternate better solution.
The calculation will become easier if you first apply an isometry. First apply an isometry such that the line $L$ will be the $x$-axis (i.e. $L: y=0$) and such that the point $A$ lies on the $y$-axis (i.e. $A=(0,q)$). I also think you may apply a dilation, so in fact you can assume that $A=(0,1)$.