The area of the triangle formed by joining the origin to the points of intersection of the line $x\sqrt{5} + 2y = 3\sqrt{5}$ and the circle $x^2 + y^2 = 10$ is

Two tangents $PQ$ and $PR$ are drawn to the circle $x^2 + y^2 - 2x - 4y - 20 = 0$ from the point $P(16, 7)$. If the centre of the circle is $C$,then the area of quadrilateral $PQCR$ is ............ $sq. \text{ units}$.

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:

$A$ 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:

If $y=mx+c$ is a common tangent to the parabola $y^2=4\sqrt{k}x$ and the circle $2x^2+2y^2=k$,then the product of the slopes of such common tangents is

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the coordinate axes cut at right angles,then: