The length of the chord intercepted by the circle $x^2+y^2-6x+8y-5=0$ on the line $2x-y=5$ is equal to $L$ units. Find $L$.

Let $S$ be the circumcircle of the triangle formed by the line $x-2y-4=0$ with the coordinate axes. If $P(-2, -4)$ is a point in the plane of the circle $S$ and $Q$ is a point on $S$ such that the distance between $P$ and $Q$ is the least,then $PQ=$

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

$A$ tangent drawn from the point $(4, 0)$ to the circle $x^2 + y^2 = 8$ touches it at a point $A$ in the first quadrant. The coordinates of another point $B$ on the circle such that the length of the chord $AB$ is $4$ are:

$A$ line meets the circle $x^2+y^2-4x-4y-8=0$ at two points $A$ and $B$. If $P(2,-2)$ is a point on the circle such that $PA=PB=2$,then the equation of the line $AB$ is:

If the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 - 2gx + g^2 - b^2 = 0$ touch each other externally,then: