Question

The locus of the point of intersection of the lines $x\sin \theta +(1-\cos \theta )y=a\sin \theta$ and $x\sin \theta -(1+\cos \theta )y+a\sin \theta =0$ is

• A. $x^2-y^2=a^2$
• B. $x^2+y^2=a^2$
• C. $y^2=ax$
• D. none of these

Answer 1

Answer: (B)

$x^2+y^2=a^2$

We have,
$x\sin \theta +(1-\cos \theta )y=a\sin \theta$ .......(1)
& $x\sin \theta -(1+\cos \theta )y+a\sin \theta =0$ ......(2)
By $\left(1\right)-\left(2\right)$ we get, $2(1-\cos \theta )y=2a\sin \theta$
$\Rightarrow y=\frac{a\sin \theta }{1-\cos \theta }$
Adding (1) and (2), we get

$2x\sin \theta -2\cos \theta y=0$
$\Rightarrow y=x\tan \theta$

$\Rightarrow x=\frac{a\cos \theta }{1-\cos \theta }$
Eliminate $\theta$ from $x$ and $y$ and get the locus.