$A$ line drawn through the point $A(5,7)$ cuts the circle $x^2+y^2-36=0$ at the points $P$ and $Q$. Then,$AP \cdot AQ=$

Let a circle $C_1$ be obtained by rolling the circle $x^2+y^2-4x-6y+11=0$ upwards $4$ units along the tangent $T$ to it at the point $(3,2)$. Let $C_2$ be the image of $C_1$ in $T$. Let $A$ and $B$ be the centers of circles $C_1$ and $C_2$ respectively,and $M$ and $N$ be respectively the feet of perpendiculars drawn from $A$ and $B$ on the $x$-axis. Then the area of the trapezium $AMNB$ is:

Let $l > 0$ be a real number,$C$ denote a circle with circumference $l$,and $T$ denote a triangle with perimeter $l$. Then:

Let $\theta$ be the angle between the circles $S \equiv x^2+y^2+2x-2y+c=0$ and $S' \equiv x^2+y^2-6x-8y+9=0$. If $c$ is an integer and $\cos \theta = \frac{5}{16}$,then the radius of the circle $S=0$ is

Two tangents are drawn from the point $P(-1, 1)$ to the circle $x^{2}+y^{2}-2x-6y+6=0$. If these tangents touch the circle at points $A$ and $B$,and if $D$ is a point on the circle such that the lengths of the segments $AB$ and $AD$ are equal,then the area of the triangle $ABD$ is equal to:

Find the equation of the chord of the circle $x^2 + y^2 - 6x + 8y = 0$ which is bisected at the point $(5, -3)$.