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 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

If from any point on the circle $x^2+y^2+2gx+2fy+c=0$,tangents are drawn to the circle $x^2+y^2+2gx+2fy+c \sin^2 \alpha + (g^2+f^2) \cos^2 \alpha = 0$,where $0 < \alpha < \frac{\pi}{2}$,then the angle between those tangents is

The shortest distance between the curves $y^2=8x$ and $x^2+y^2+12y+35=0$ is:

For the given circle $2x^2 + 2y^2 = 5$ and the parabola $y^2 = 4\sqrt{5}x$:
Statement-$I$: The equation of the common tangent to these curves is $y = x + \sqrt{5}$.
Statement-$II$: If the line $y = mx + \frac{\sqrt{5}}{m} (m \neq 0)$ is a common tangent,then $m$ satisfies $m^4 - 3m^2 + 2 = 0$.

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the coordinate axes cut at right angles,then: