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:

Let $a$ and $b$ be non-zero real numbers. Then,the equation $(a x^2+b y^2+c)(x^2-5 x y+6 y^2)=0$ represents

Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16x$. If one of these tangents touches the two curves at $Q$ and $R$,then $(QR)^2$ is equal to

$A$ 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:

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines $x + (k-1)y + 3 = 0$ and $2x + ky - 4 = 0$. If the line $x - y + 2 = 0$ intersects the circle at the points $A$ and $B$, then $(AB)^2$ is equal to:

Let the point $P$ be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point $P$ intersects the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the points $R$ and $S$, then the maximum value of $(PR + PS)^2$ is: