2015/084) Prove that the locus of the center of the circle

$\frac{1}{2}(x^2 + y^2) + x \cos(\theta) + y \sin(\theta) - 4 = 0$ is $x^2 + y^2 = 1$

Solution

we have $(x^2 + 2 x \cos \theta) + (y^2 + 2y \sin \theta ) = 8$
or $(x^2 + 2 x \cos \theta+1) + (y^2 + 2y \sin \theta+ 1 ) = 10$

or $( x + \cos \theta)^2 + (y + \sin \theta ) ^2 = 10$
So the locus of the centre
$x = - \cos \theta$
$y = - \sin \theta$
to eliminate $\theta$
or $x^2 + y ^2 = 1$
above is the locus
so above is true