The locus of the centre of a circle which cuts orthogonally the circle $x^2 + y^2 - 20x + 4 = 0$ and which touches the line $x = 2$ is:

Find the locus of a point such that the sum of the squares of its distances from the axes is $4$.

Let $\Gamma$ be a circle with diameter $AB$ and centre $O$. Let $l$ be the tangent to $\Gamma$ at $B$. For each point $M$ on $\Gamma$ different from $A$,consider the tangent $t$ at $M$ and let it intersect $l$ at $P$. Draw a line parallel to $AB$ through $P$ intersecting $OM$ at $Q$. The locus of $Q$ as $M$ varies over $\Gamma$ is

The locus of the centre of the circle $\frac{1}{2} (x^2 + y^2) + x \cos \theta + y \sin \theta - 4 = 0$ is :-

$A$ point moves such that the sum of the squares of its distances from the points $(1, 2)$ and $(-2, 1)$ is always $6$. Then, its locus is

The angle between a pair of tangents drawn from a point $P$ to the circle ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$ is $2\alpha$. The equation of the locus of the point $P$ is