$A$ and $B$ are fixed points with coordinates $(0, a)$ and $(0, b)$ where $a > 0$ and $b > 0$. $P$ is a variable point $(x, 0)$ on the $x$-axis. If the angle $\angle APB$ is maximum, then:

If $A=(1,2)$,$B=(2,1)$ and $P$ is any point satisfying the condition $PA+PB=3$,then the equation of the locus of $P$ 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:

What does the equation $\sqrt{(x - 2)^2 + (y + 1)^2} + \sqrt{(x + 2)^2 + (y + 4)^2} = 5$ represent?

The locus of the point which is equidistant from the point $(1,1)$ and the line $x+y+1=0$ is

Find the locus of point $P$ such that $2PA = 3PB$,where $A = (0, 0)$ and $B = (4, -3)$.