The minimum distance between the curves $x^2 + y^2 + 4x + 16y + 66 = 0$ and $y^2 = 8x$ is:

Prove that the curves $y^{2}=4x$ and $x^{2}+y^{2}-6x+1=0$ touch each other at the point $(1,2)$.

The equation of the common tangent to the parabola $y^2=8x$ and the circle $x^2+y^2=2$ is $ax+by+2=0$. If $-\frac{a}{b} > 0$,then $3a^2+2b+1=$

Let circle $C$ be the image of $x^2+y^2-2x+4y-4=0$ in the line $2x-3y+5=0$. Let $A$ be the point on $C$ such that $OA$ is parallel to the $x$-axis and $A$ lies on the right hand side of the centre $O$ of $C$. If $B(\alpha, \beta)$,with $\beta < 4$,lies on $C$ such that the length of the arc $AB$ is $(1/6)^{\text{th}}$ of the perimeter of $C$,then $\beta - \sqrt{3}\alpha$ is equal to

An equilateral triangle $OAB$ is inscribed in the parabola $y^{2}=4x$ with the vertex $O$ at the origin. Then the minimum distance of the circle having $AB$ as a diameter from the origin is:

The area (in $sq. units$) of the smaller of the two circles that touch the parabola $y^2 = 4x$ at the point $(1, 2)$ and the $x$-axis is