Question

Find the locus of the point of intersection of line $x\cos \alpha +y\sin \alpha =a$

and $x\sin \alpha -y\cos \alpha =b$.

Answer 1

Consider the equation

$x\cos \alpha +y\sin \alpha =a$

$x\sin \alpha -y\cos \alpha =b$

Let $P(h,k)$ be the point satisfying the equation then the above equation becomes

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

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

squaring and adding eqns$\left(1\right)$ and $\left(2\right)$ we get

$h^2{\cos }^2\alpha +k^2{\sin }^2\alpha +2hk\sin \alpha \cos \alpha =a^2$

$h^2{\sin }^2\alpha +k^2{\cos }^2\alpha -2hk\sin \alpha \cos \alpha =b^2$

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

$\Rightarrow h^2+k^2=a^2+b^2$ since ${\cos }^2\alpha +{\sin }^2\alpha =1$

Replace $h\to x$ and $k\to y$ we get equation of the locus as

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