Question

Three distinct unit circles are given such that each circle touches the other two. The radius of another circle which touches all the given circles can be

• A. $\sqrt{3}-1$
• B. $\sqrt{3}+1$
• C. $\frac{2\sqrt{3}}{3}-1$
• D. $\frac{2\sqrt{3}}{3}+1$

Answer 1

Answer: (C), (D)

$\frac{2\sqrt{3}}{3}+1$
$\frac{2\sqrt{3}}{3}-1$

The radius of the circles is $1$
Hence, the length of sides of the equilateral triangle formed by joining the centres of the circles will be $2$.
Now, the radius of the other(bigger) circle would be
$1+OC$ ($O$ is the center of the circle)
$\angle OCD={30}^0$ ...(OC is the angle bisector).
Now, if we construct a perpendicular to the side BC from the centre of the larger circle, $OD=BD=1unit$
Hence,
$\frac{1}{OC}=\cos \left({30}^0\right)$
$\frac{1}{OC}=\frac{\sqrt{3}}{2}$
$OC=\frac{2}{\sqrt{3}}$
$=\frac{2\sqrt{3}}{3}$
Now, the radius of the bigger circle will be $\frac{2\sqrt{3}}{3}+1$
However, a smaller circle can also be drawn, concentric with the larger circle and touching all the three circle, in the space in between the three circles, (Refer to the figure.)
The radius of that circle will be $\frac{2\sqrt{3}}{3}-1$.