Question

Prove that the locus of the centre of the circle $\frac{1}{2}(x^2+y^2)+xcos\theta +ysin\theta -4=0$ is $x^2+y^2=1$.

Answer 1

Multiplying the relation throughout with 2,

$x^2+y^2+2x\cos \theta +2y\sin \theta +1=9\Rightarrow (x+cos\theta {)}^2+(y+\sin \theta {)}^2=3^2$

ie, The centre of the circle is $(\cos \theta ,\sin \theta )$

So, Inorder to find locus,

Let $x=\cos \theta$ and $y=\sin \theta$

$\Rightarrow {\cos }^2\theta +{\sin }^2\theta =1\Rightarrow x^2+y^2=1$