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

The locus of the point of intersection of tangents to the circle $x = a \cos \theta, y = a \sin \theta$ at the points whose parametric angles differ by $\pi / 2$ is:

Let $x$ and $y$ be real numbers satisfying the equation $x^2 - 4x + y^2 + 3 = 0$. If the maximum and minimum values of $x^2 + y^2$ are $M$ and $m$ respectively,then the numerical value of $M - m$ is:

$A$ circle touches the $X$-axis and also touches the circle with radius $2$ and center at $(0, 3)$. Find the locus of the center of the circle.

$A$ point $P(x, y)$ is such that the sum of squares of its distances from the coordinate axes is equal to the square of its distance from the line $x-y=1$. Then the equation of the locus of $P$ is

If $P(x_1, y_1)$ is a point such that the lengths of the tangents from it to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ are in the ratio $2:3$,then the locus of $P$ is