The line $x+y=k$ meets the curve $x^2+y^2-2x-4y+2=0$ at two points $A$ and $B$. If $O$ is the origin and $\angle AOB=90^{\circ}$,then the value of $k$ $(k>1)$ is

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:

The equation of a circle which touches the straight lines $x+y=2$ and $x-y=2$ and also touches the circle $x^2+y^2=1$ is:

The equations of the circles which touch the lines $3x - 4y + 1 = 0$ and $4x + 3y - 7 = 0$ and pass through the point $(2, 3)$ are:

Consider the two curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Then,

$A$ line drawn through the point $P(4, 7)$ cuts the circle $x^2 + y^2 = 9$ at the points $A$ and $B$. Then $PA \cdot PB$ is equal to