The locus of the point of intersection of perpendicular tangents to the circle $x^{2}+y^{2}=16$ is

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.

If ${\theta _1}$ and ${\theta _2}$ are the inclinations of the tangents drawn from a point $P(h, k)$ to the circle ${x^2} + {y^2} = {a^2}$ with the $x$-axis,then the locus of $P$,given that $\cot {\theta _1} + \cot {\theta _2} = c$,is:

The locus of the centre of the circle touching the $x$-axis and passing through the point $(-1, 1)$ is

If $A(2, 3)$ and $B(3, -2)$ are two fixed points and $P(x, y)$ is a variable point satisfying the condition $|PA - PB| = 2$,then the locus of $P$ is

$A$ point $P$ moves such that the distance from $(0,2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1,0)$. Then the locus of the point is