From three non-collinear points,we can draw:

Let the centre of the circle be in the first quadrant and lie on the line $2x - y = 4$. Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is . . . . . .

The point $(5, -7)$ lies outside the circle:

Let $B$ be the centre of the circle $x^{2}+y^{2}-2x+4y+1=0$. Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A(3,1)$. Then $8 \left(\frac{\text{Area } \triangle APQ}{\text{Area } \triangle BPQ}\right)$ is equal to:

The length of the transverse common tangent of the circles $x^2+y^2-2x+4y+4=0$ and $x^2+y^2+4x-2y+1=0$ is

If the shortest distance from $(2,-14)$ to the circle $x^2+y^2+6x+4y-12=0$ is $d$ and the length of the tangent drawn from the same point to the circle is $l$,then $\sqrt{d+l}=$