Question

Whatever be the value of $\alpha$, prove that the locus of the intersection of the straight lines $x\cos \alpha +y\sin \alpha =a$ and $x\sin \alpha -y\cos \alpha =b$ is a circle.

Answer 1

Given equations are $x\cos \alpha +y\sin \alpha =a$ ... $\left(i\right)$ and $x\sin \alpha -y\cos \alpha =b$ ... $\left(ii\right)$

Square and add the both the equations, we get

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

$\Rightarrow (x^2{\cos }^2\alpha +y^2{\sin }^2\alpha +2xy\sin \alpha \cos \alpha )+(x^2{\sin }^2\alpha +y^2{\cos }^2\alpha -2xy\sin \alpha \cos \alpha )=a^2+b^2$

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

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

Clearly above equation represents a circle.