The length of the chord joining the points $(4 \cos \theta, 4 \sin \theta)$ and $(4 \cos (\theta+60^{\circ}), 4 \sin (\theta+60^{\circ}))$ of the circle $x^{2}+y^{2}=16$ is

Given three collinear points $A(3,1)$,$B(7,-1)$,and $C(5,0)$. The length of a tangent drawn from $A$ to any circle that passes through $B$ and $C$ is ....... units.

Let $S_1, S_2,$ and $S_3$ be three circles of unit radius which touch each other externally. The common tangents to each pair of circles are drawn and extended so that they intersect and form a triangle $ABC$ with circumradius $R$. Then $R$ is equal to

If a circle $S$ with radius $5$ touches the circle $x^2+y^2-6x-4y-12=0$ at $(-1,-1)$,then the length of the tangent from the centre of the circle $S$ to the given circle is

Let $y=x$ be the equation of a chord of the circle $C_{1}$ (in the closed half-plane $x \ge 0$) of diameter $10$ passing through the origin. Let $C_{2}$ be another circle described on the given chord as its diameter. If the equation of the chord of the circle $C_{2}$, which passes through the point $(2, 3)$ and is farthest from the center of $C_{2}$, is $x+ay+b=0$, then $a-b$ is equal to:

The equation of the circumcircle of the triangle formed by the lines $y + \sqrt{3}x = 6$,$y - \sqrt{3}x = 6$,and $y = 0$ is: