The point of contact of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2-2x+2y+1=0$ is

If $(a \cos \theta_i, a \sin \theta_i)$ for $i = 1, 2, 3$ represent the vertices of an equilateral triangle inscribed in a circle $x^2 + y^2 = a^2$,then:

The angle subtended by the chord $x+y-1=0$ of the circle $x^2+y^2=1$ at the origin is

Let a circle $C$ touch the lines $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$,where $K_{1}, K_{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(-1, 2)$ and $L_{2}$ at $(3, -6)$,then the equation of the circle $C$ is:

If the point $(2 \cos \theta, 2 \sin \theta)$ for $\theta \in (0, 2 \pi)$ lies in the region between the lines $x+y=2$ and $x-y=2$ containing the origin, then $\theta$ lies in

$x^2+y^2+2x-6y-6=0$ and $x^2+y^2-6x-2y+k=0$ are two intersecting circles and $k$ is not an integer. If $\theta$ is the angle between the two circles and $\cos \theta = \frac{-5}{24}$,then $k=$