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

Let the parabola $y=x^2+px-3$ meet the coordinate axes at the points $P, Q$ and $R$. If the circle $C$ with center at $(-1,-1)$ passes through the points $P, Q$ and $R$,then the area of $\triangle PQR$ is:

For different real non-zero numbers $x_1, x_2, x_3$ and $x_4$,suppose the points $(x_1, \frac{1}{x_1}), (x_2, \frac{1}{x_2}), (x_3, \frac{1}{x_3})$ and $(x_4, \frac{1}{x_4})$ lie on the boundary of a circle of radius $4$. Then,the value of $x_1 x_2 x_3 x_4$ is

Let $C$ be the circle of minimum area touching the parabola $y=6-x^2$ and the lines $y=\sqrt{3}|x|$. Then,which one of the following points lies on the circle $C$?

If $(\alpha, \beta)$ is a point on the circle whose centre is on the $x$-axis and which touches the line $x + y = 0$ at $(2, -2)$,then the greatest value of $\alpha$ is

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