If one end of a diameter of the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is $(1, 0)$,then the other end of the diameter is:

Find the equation of the pair of straight lines parallel to the $x$-axis and touching the circle $x^2 + y^2 - 6x - 4y - 12 = 0$.

If $P(0,0), Q(1,0)$ and $R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ are three given points, then the centre of the circle for which the lines $PQ, QR$ and $RP$ are the tangents is

The circle $x^2 + y^2 - 4x - 8y - 5 = 0$ will intersect the line $3x - 4y = m$ in two distinct points,if:

If the angle between the circles $x^2+y^2-2x-4y+c=0$ and $x^2+y^2-4x-2y+4=0$ is $60^{\circ}$,then $c=$

$A$ circle passing through the point $P(\alpha, \beta)$ in the first quadrant touches the two coordinate axes at the points $A$ and $B$. The point $P$ is above the line $AB$. The point $Q$ on the line segment $AB$ is the foot of the perpendicular from $P$ on $AB$. If $PQ$ is equal to $11$ units,then the value of $\alpha \beta$ is $.............$.