Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16x$. If one of these tangents touches the two curves at $Q$ and $R$,then $(QR)^2$ is equal to

Let $a$ and $b$ be two non-zero real numbers. The equation $(ax^2 + by^2 + c)(x^2 - 5xy + 6y^2) = 0$ represents:

If the tangent at the point $\left(4 \cos 2 \theta, \frac{16}{\sqrt{11}} \sin 2 \theta\right)$ on the ellipse $16 x^2+11 y^2=256$ touches the circle $x^2+y^2-2 x=15$,then $\theta=$

If any tangent drawn to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ touches one of the circles $x^2 + y^2 = \alpha^2$,then the range of $\alpha$ is

The common chord of two intersecting circles $c_1$ and $c_2$ subtends angles of $90^\circ$ and $60^\circ$ at their respective centres. If the distance between their centres is $\sqrt{3} + 1$,then the radii of $c_1$ and $c_2$ are:

For $n \in N$,let $S_{n} = \{ z \in C : |z - 3 + 2i| = \frac{n}{4} \}$ and $T_{n} = \{ z \in C : |z - 2 + 3i| = \frac{1}{n} \}$. Then the number of elements in the set ${ n \in N : S_{n} \cap T_{n} = \phi }$ is.