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 $PQ=$

Two circles with equal radii intersect at the points $(0, 1)$ and $(0, -1).$ The tangent at the point $(0, 1)$ to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is

Three circles of radii $a, b, c$ $(a < b < c)$ touch each other externally. If they have the $x$-axis as a common tangent,then:

Let $ABCD$ be a trapezium with parallel sides $AB$ and $CD$ such that the circle $S$ with $AB$ as its diameter touches $CD$. Further,the circle $S$ passes through the mid-points of the diagonals $AC$ and $BD$ of the trapezium. The smallest angle of the trapezium is

Let $C_1$ and $C_2$ be two circles touching each other externally at point $A$. Let $AB$ be the diameter of circle $C_1$. Draw a secant $BA_3$ to circle $C_2$,intersecting circle $C_1$ at point $A_1$ (where $A_1 \neq A$) and circle $C_2$ at points $A_2$ and $A_3$. If $BA_1 = 2$,$BA_2 = 3$,and $BA_3 = 4$,then the radii of circles $C_1$ and $C_2$ are respectively:

Find the maximum distance of the point $K(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$.