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 $C$?

The line $x=2y$ intersects the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at the points $P$ and $Q$. The equation of the circle with $PQ$ as diameter is

The angle of intersection between the curves $y^2+x^2=a^2 \sqrt{2}$ and $x^2-y^2=a^2$ is

The circle ${x^2} + {y^2} = 4$ cuts the line joining the points $A(1, 0)$ and $B(3, 4)$ in two points $P$ and $Q$. Let $\frac{BP}{PA} = \alpha$ and $\frac{BQ}{QA} = \beta$. Then $\alpha$ and $\beta$ are roots of the quadratic equation

The equation of a circle touching the parabola $y = x^2$ at the point $(1, 1)$ and passing through the point $(2, 2)$ is:

Let the line $L: \sqrt{2}x + y = \alpha$ pass through the point of intersection $P$ (in the first quadrant) of the circle $x^2 + y^2 = 3$ and the parabola $x^2 = 2y$. Let the line $L$ touch two circles $C_1$ and $C_2$ of equal radius $2\sqrt{3}$. If the centres $Q_1$ and $Q_2$ of the circles $C_1$ and $C_2$ lie on the $y$-axis,then the square of the area of the triangle $PQ_1Q_2$ is equal to: