Two circles of radii $4$ cms $\&\,\, 1\,\, cm$ touch each other externally and $\theta$ is the angle contained by their direct common tangents. Then $sin \theta =$
$\frac{{{\rm{24}}}}{{{\rm{25}}}}$
$\frac{{{\rm{12}}}}{{{\rm{25}}}}$
$\frac{{\rm{3}}}{{\rm{4}}}$
none
Let $r_{1}$ and $r_{2}$ be the radii of the largest and smallest circles, respectively, which pass through the point $(-4,1)$ and having their centres on the circumference of the circle $x^{2}+y^{2}+2 x+4 y-4= 0.$ If $\frac{r_{1}}{r_{2}}=a+b \sqrt{2}$, then $a+b$ is equal to:
The equation of radical axis of the circles ${x^2} + {y^2} + x - y + 2 = 0$ and $3{x^2} + 3{y^2} - 4x - 12 = 0,$ is
Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals
The equation of the circle having its centre on the line $x + 2y - 3 = 0$ and passing through the points of intersection of the circles ${x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${x^2} + {y^2} - 4x - 2y + 4 = 0$, 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)