If a circle $C,$ whose radius is $3,$ touches externally the circle, $x^2 + y^2 + 2x - 4y - 4 = 0$ at the point $(2, 2),$ then the length of the intercept cut by circle $c,$ on the $x-$ axis is equal to
$\sqrt 5$
$2\sqrt 3$
$3\sqrt 2$
$2\sqrt 5$
The two circles ${x^2} + {y^2} - 2x + 6y + 6 = 0$ and ${x^2} + {y^2} - 5x + 6y + 15 = 0$ touch each other. The equation of their common tangent is
In the figure shown, radius of circle $C_1$ be $ r$ and that of $C_2$ be $\frac{r}{2}$ , where $r= \frac {1}{3} PQ,$ then length of $AB$ is (where $P$ and $Q$ being centres of $C_1$ $\&$ $C_2$ respectively)
If a circle $C$ passing through $(4, 0)$ touches the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ externally at a point $(1, -1),$ then the radius of the circle $C$ is
Two circles of radii $4$ cms $\&\,\, 1\,\, cm$ touch each other externally and $\theta$ is the angle contained by their direct common tangents. Then $sin \theta =$
The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be