If a circle passes through the point $(a, b)$ and cuts the circle ${x^2} + {y^2} = 4$ orthogonally,then the locus of its centre 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:

Two perpendicular tangents to the circle $x^2 + y^2 = a^2$ meet at point $P$. The equation of the locus of $P$ is:

If one end of the diameter is $(1, 1)$ and the other end lies on the line $x+y=3$,then the locus of the centre of the circle 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?

The locus of the centre of the circles passing through the origin and cutting off a chord of length $2$ units on the line $x=1$ is