The locus of a point which is at a distance of $4$ units from $(3, -2)$ in the $xy$-plane is

Let the circle $x^{2}+y^{2}=4$ intersect the $x$-axis at the points $A(a,0), a>0$ and $B(b,0)$. Let $P(2 \cos \alpha, 2 \sin \alpha), 0 < \alpha < \frac{\pi}{2}$ and $Q(2 \cos \beta, 2 \sin \beta)$ be two points such that $(\alpha - \beta) = \frac{\pi}{2}$. Then the point of intersection of $AQ$ and $BP$ lies on:

$A$ variable circle passes through the fixed point $(2,0)$ and touches the $y$-axis. Then the locus of its centre is

Two rods of lengths $a$ and $b$ slide along the coordinate axes in such a way that their ends are always concyclic. The locus of the center of the circle passing through the ends is:

$A$ circle passes through the point $(3, 4)$ and cuts the circle $x^2 + y^2 = a^2$ orthogonally. The locus of its centre is a straight line. If the distance of this straight line from the origin is $25$,then $a^2$ is equal to:

The image of every point lying on the curve $x^2+y^2=1$ in the line $x+y=1$ satisfies the equation: