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

The power of the point $B(-1, 1)$ with respect to the circle $S \equiv x^2+y^2-2x-4y+3=0$ is $p$. If the length of the tangent drawn from $B$ to the circle $S=0$ is $t$,then the point $(2, 3)$ with respect to the circle $S^{\prime}=0$ having centre at $(p, t^2)$ and passing through the origin:

$A$ variable line $ax + by + c = 0$,where $a, b, c$ are in $A.P.$,is normal to a circle $(x - \alpha)^2 + (y - \beta)^2 = \gamma$,which is orthogonal to the circle $x^2 + y^2 - 4x - 4y - 1 = 0$. The value of $\alpha + \beta + \gamma$ is equal to

$P$ 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:

If the lines $kx + 2y - 4 = 0$ and $5x - 2y - 4 = 0$ are conjugate with respect to the circle $x^2 + y^2 - 2x - 2y + 1 = 0$,then $k$ is equal to

The equation of a common tangent to the circle $x^2+y^2=16$ and the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ is