Given two fixed points $A(-2, 1)$ and $B(3, 0)$,find the locus of a point $P$ which moves such that the angle $\angle APB$ is always a right angle.

The locus of the centre of a circle passing through $(a, b)$ and cutting orthogonally to the circle $x^2 + y^2 = p^2$ is

If $A$ and $B$ are two fixed points in a plane and $P$ is another variable point such that $PA^2 + PB^2 = k$ (where $k$ is a constant),then the locus of the point $P$ is:

If $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ touches the circle $x^2 + y^2 = a^2$,then the point $(\frac{1}{\alpha}, \frac{1}{\beta})$ lies on a/an

If $P(x, y)$ is a point such that the ratio of the squares of the lengths of the tangents from $P$ to the circles $x^2 + y^2 + 2x - 4y - 20 = 0$ and $x^2 + y^2 - 4x + 2y - 44 = 0$ is $2 : 3$,then the locus of $P$ is a circle with centre

The locus of the point of intersection of perpendicular tangents drawn to the circle $x^2+y^2=10$ is