If $P(1, 0)$,$Q(-1, 0)$,and $R(2, 0)$ are three given points,then find the locus of $S(x, y)$ which satisfies the relation $SQ^2 + SR^2 = 2SP^2$.

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:

The locus of the third vertex of a right-angled triangle,the ends of whose hypotenuse are $(1,2)$ and $(4,5)$,is

If the tangents drawn from a point $P$ to the circle $x^2 + y^2 = a^2$ are perpendicular to the tangents drawn to the circle $x^2 + y^2 = b^2$,then the locus of $P$ is:

At which point on the line $x = 3$ are the tangents drawn to the circle $x^2 + y^2 = 8$ perpendicular to each other?

If a circle passes through the point $(1, 2)$ and intersects the circle $x^2 + y^2 = 4$ orthogonally,then the equation of the locus of its center is: