Question

The equation of the locus of the point of intersection of the straight 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. $y=\pm ax$
• B. $x=\pm ay$
• C. $y^2=4x$
• D. $x^2+y^2=a^2$

Answer 1

Answer: (D)

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

Given equations are
$x\sin \theta +(1-\cos \theta )$ $y=a\sin \theta$ and $x\sin \theta -(1-\cos \theta )y+a\sin \theta =0$
On subtracting, we get
$2y=2a\sin \theta \Rightarrow y=a\sin \theta$
and $x=a\cos \theta$
$\therefore x^2+y^2=a^2{\cos }^2\theta +a^2{\sin }^2\theta =a^2$