If $\theta$ is the angle made by the common tangent to the circle $x^2+y^2=16$ and the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ with the positive $X$-axis,then $\cos 2 \theta=$

$A$ circle $C$ of radius $2$ lies in the second quadrant and touches both the coordinate axes. Let $r$ be the radius of a circle that has its centre at the point $(2, 5)$ and intersects the circle $C$ at exactly two points. If the set of all possible values of $r$ is the interval $(\alpha, \beta)$,then $3 \beta - 2 \alpha$ is equal to:

Let $\theta$ be the acute angle between the tangents to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ at their point of intersection in the first quadrant. Then $\tan \theta$ is equal to:

Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If $(1, \alpha)$ lies on $C$,then $10 \alpha^2$ is equal to $.........$

The power of the point $B(-1, 1)$ with respect to the circle $S \equiv x^2+y^2-2x-4y+3=0$ is $p$. If the length of the tangent drawn from $B$ to the circle $S=0$ is $t$,then the point $(2, 3)$ with respect to the circle $S^{\prime}=0$ having centre at $(p, t^2)$ and passing through the origin:

$A$ variable circle passes through the fixed point $A(p, q)$ and touches the $x$-axis. The locus of the other end of the diameter through $A$ is