Let $A=(4,0)$ and $B=(0,12)$ be two points in the plane. The locus of a point $C(x, y)$ such that the area of $\triangle ABC$ is $18$ sq units is

$A$ variable line $L$ passing through the origin cuts two parallel lines $x-y+10=0$ and $x-y+20=0$ at two points $A$ and $B$ respectively. If $P$ is a point on line $L$ such that $OA, OP, OB$ are in harmonic progression,then the locus of $P$ is

The equation of the locus of the foot of the perpendicular drawn from the origin to a line passing through a fixed point $(a, b)$ is:

The locus of $P$ such that the area of $\Delta PAB = 12 \text{ sq. units}$,where $A(2, 3)$ and $B(-4, 5)$ is:

If $A=(5,3)$,$B=(3,-2)$ and a point $P$ is such that the area of the triangle $PAB$ is $9$,then the locus of $P$ represents

If for a variable line $\frac{x}{a} + \frac{y}{b} = 1$,the condition $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$ ($c$ is a constant) is satisfied,then the locus of the foot of the perpendicular drawn from the origin to the line is: