Question

A variable circle passes through the point of intersection of any two fixed straight lines of cuts off from them portions $OA$ and $OB$ such that $pOA+qOB=1$. Show that this circle always passes through a fixed point.

Answer 1

Choose one fixed line along x-axis and the other through origin as $y=mx$ where $m=\tan \theta (\theta$ fixed as lines are fixed).
Let $OA=a\therefore A$ is $(a,0)$ and $OB=r$
$\therefore B$ is $(r\cos \theta ,r\sin \theta )$
We are given that
$p.OA.q.OB=1$
o $pa+qr=1.....\left(1\right)$
where $a$ and $r$ are variables
Now let the circel $OAB$ be chosen as
$x^2+y^2+2gx+2fy=0.....\left(2\right)$
(Mind $c=0$ as it passes through $O(0,0)$. Note that $A$ is on x-axis but not $B$). It passes through $A(a,0)$.
$\therefore a^2+2ga=0\therefore 2g=-a$
The point $B$ is $(r\cos \theta ,r\sin \theta )$ which lies on the circle.
$\therefore r^2+2r(g\cos \theta +f\sin \theta )=0$
$\therefore r=-2(g\cos \theta +f\sin \theta )$
$\therefore r=a\cos \theta -2f\sin \theta \because 2g=-a$
Putting in $\left(1\right)$, we get
$pa+q(a\cos \theta -2f\sin \theta )=1$
$\therefore (p+q\cos \theta )a-1=2fq\sin \theta$
Putting the values of $g$ and $f$ in $\left(2\right)$ the equation of circle becomes
$x^2+y^2-ax+\frac{\left[\right(p+q\cos \theta )a-1]}{q\sin \theta }y=0$
Collect the terms of variable $a$.
$(x^2+y^2-\frac{y}{q\sin \theta })+a(\frac{p+q\cos \theta }{q\sin \theta }y-x)$
It is of the form $S+\lambda P=0$, where $S=0$ is a circle and $P=0$ is the equation of line. It clearly passes through the points of intersection of $S$ and $P$ which are fixed as their equations contain all constants.