The locus of the point of intersection of the tangents to the circle $x^2+y^2=16$,such that the angle between them is $60^{\circ}$,is

Let $A(5,4)$ and $B(5,-4)$ be two points. If $P(x,y)$ is a point in the coordinate plane such that $\angle APB = \frac{\pi}{4}$,then the point $P$ lies on the curve

Let $\Gamma$ be a circle with diameter $AB$ and centre $O$. Let $l$ be the tangent to $\Gamma$ at $B$. For each point $M$ on $\Gamma$ different from $A$,consider the tangent $t$ at $M$ and let it intersect $l$ at $P$. Draw a line parallel to $AB$ through $P$ intersecting $OM$ at $Q$. The locus of $Q$ as $M$ varies over $\Gamma$ is

$A$ circle of constant radius $a$ passes through the origin $O$ and cuts the coordinate axes at points $P$ and $Q$. The equation of the locus of the foot of the perpendicular from $O$ to $PQ$ is:

The locus of the point whose ratio of distance from the origin to its distance from $(-2, -3)$ is $5: 7$,is given by:

The locus of the point given by the equations $x = \frac{2at}{1 + t^2}$ and $y = \frac{a(1 - t^2)}{1 + t^2}$ for $-1 \le t \le 1$ is a