Question

$O$ is a fixed point and $AP$ and $BQ$ are two fixed parallel straight lines. $BOA$ is perpendicular to both and $\angle POQ$ is a right angle. Prove that the locus of the foot of the perpendicular from $O$ on $PQ$ is a circle whose diameter is $AB$.

Answer 1

We take $O$ as origin and x-axis along $BOA$. Let the coordinates of $A$ and $B$ be $(a,0)$ and $(-b,0)$. Let the fixed parallel lines $AP$ and $BQ$ be taken parallel to y-axis so that $BOA$ is perpendicular to both.

If $\angle POA=\theta$ then as $\angle POQ={90}^o$

We have $\angle QOB={90}^o-\theta$

Coordinates of $P$ are $(a,a\tan \theta )$ and those of $Q$ are

$(-b,b\cot \theta )$

Equation to line $PQ$ is

$y-a\tan \theta =\frac{b\cot \theta -a\tan \theta }{-(b+a)}(x-a)$

or $x(b\cot \theta -a\tan \theta )+y(a+b)=ab(\cot \theta +\tan \theta )$

or changing into $\sin \theta$ and $\cos \theta$ it becomes

$x(b{\cos }^2\theta -a{\sin }^2\theta )+y(a+b)\sin \theta \cos \theta =ab$

Multiplying by $2$, it becomes

$x(2b{\cos }^2\theta -2a{\sin }^2\theta )+y(a+b)\sin 2\theta =2ab....\left(1\right)$

Also equation to perpendicular $OR$ on $PQ$ is

$x(a+b)\sin 2\theta -y(2b{\cos }^2\theta -2a{\sin }^2\theta )=\mathrm{0....}\left(2\right)$

But $(2b{\cos }^2\theta -2a{\sin }^2\theta )=b(1+\cos 2\theta )-a(1+\cos 2\theta )=(a+b)1+\cos 2\theta -(a-b)$

Hence the two lines (1) and (2) can be written as

$x(a+b)\cos 2\theta +y(a+b)\sin 2\theta =2ab+(a-b)x.....\left(3\right)$

and $x(a+b)\sin 2\theta -y(a+b)\cos 2\theta =-(a-b)y.....\left(4\right)$

In order to find the locus of point of intersection $R$, we have to eliminate $\theta$ for which we square and add the equations (3) and (4)

$\therefore x^2{(a+b)}^2+y^2{(a+b)}^2=4a^2{b}^2+4ab(a-b)x+{(a-b)}^2(x^2+y^2)$

or $(x^2+y^2)({(a+b)}^2-{(a-b)}^2)=4ab(ab+(a-b\left)x\right)$

$x^2-(a-b)x-ab+y^2=0or(x-a)(y+b)+y^2=0$

above represents the equation of a circle on $AB$ as diameter.