$A$ parabola $y = ax^2 + bx + c$ crosses the $x$-axis at $(\alpha, 0)$ and $(\beta, 0)$,both to the right of the origin. $A$ circle also passes through these two points. The length of a tangent from the origin to the circle is:

Let $a$ and $b$ be two non-zero real numbers. The equation $(ax^2 + by^2 + c)(x^2 - 5xy + 6y^2) = 0$ represents:

Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then
$(A)$ $SP=2\sqrt{5}$
$(B)$ $SQ:QP=(\sqrt{5}+1):2$
$(C)$ the $x$-intercept of the normal to the parabola at $P$ is $6$
$(D)$ the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

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:

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. Find the equation of the circle with $AB$ as its diameter.

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