Question

Find the LOCUS of point of intersection of the lines
$x\cos \alpha +y\sin \alpha =a$ and $x\sin \alpha -y\cos \alpha =b$, where $\alpha$ is a parameter.

• A. $x^2-y^2=a^2+b^2$
• B. $x^2-y^2=a^2-b^2$
  
• C. $x^2+y^2=a^2-b^2$
  
• D. $x^2+y^2=a^2+b^2$
  

Answer 1

Answer: (D)

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

Let $P(h,k)$ be the point of intersection of the given lines.

At $P(h,k)$ the given equations become,

$h\cos \alpha +k\sin \alpha =a$ ...... $\left(1\right)$

$h\sin \alpha -k\cos \alpha =b$ ....... $\left(2\right)$

Squaring and adding both the equations, we get

$(h\cos \alpha +k\sin \alpha {)}^2+(h\sin \alpha -k\cos \alpha {)}^2=a^2+b^2$

$(h^2{\cos }^2\alpha +k^2{\sin }^2\alpha +2h\cos \alpha k\sin \alpha )+(h^2{\sin }^2\alpha +k^2{\cos }^2\alpha -2h\sin \alpha k\cos \alpha )=a^2+b^2$

$(h^2{\cos }^2\alpha +h^2{\sin }^2\alpha )+(k^2{\sin }^2\alpha +k^2{\cos }^2\alpha )=a^2+b^2$

$h^2({\cos }^2\alpha +{\sin }^2\alpha )+k^2({\sin }^2\alpha +{\cos }^2\alpha )=a^2+b^2$

$h^2+k^2=a^2+b^2$ .......... $(\because {\cos }^2\alpha +{\sin }^2\alpha =1)$

$\therefore x^2+y^2=a^2+b^2$ is the locus of the point of intersection of the lines given.

This is a circle with origin as center and radius $\sqrt{a^2+b^2}$