Three equal circles each of radius r touch one another. The radius of the circle which touching by all the three given circles internally is
A) $\left( {2 + \sqrt 3 } \right)r$
B) $\dfrac{{\left( {2 + \sqrt 3 } \right)}}{{\sqrt 3 }}r$
C) $\dfrac{{\left( {2 - \sqrt 3 } \right)}}{{\sqrt 3 }}r$
D) $\left( {2 - \sqrt 3 } \right)r$
VerifiedHint:
First, we’ll perform required construction according to the question then we will find the OE from $\Delta OAE$ and in the end we’ll find the radius using the equation $OB = OE + EB$.
Complete step by step solution:
Here, it is given that three equal circles each of radius r touch one another.
Now, we will find that the radius of the circle which touching by all the three given circles internally
From the figure it is clear that $\Delta DEF$ is an equilateral triangle
In $\Delta OAE$ ,
$\dfrac{{AE}}{{OE}} = \cos {30^ \circ }$
We know that $\cos {30^ \circ }\dfrac{{\sqrt 3 }}{2}$
$OE = \dfrac{{AE}}{{\cos {{30}^{^ \circ }}}}$
$OE = \dfrac{r}{{\dfrac{{\sqrt 3 }}{2}}}$
$OE = \dfrac{{2r}}{{\sqrt 3 }}$
$\because OB = OE + EB$
$\therefore OB = \dfrac{{2r}}{{\sqrt 3 }} + r$
$OB = \dfrac{{\left( {\sqrt 3 + 2} \right)r}}{{\sqrt 3 }}$
$\therefore R = \dfrac{{\left( {\sqrt 3 + 2} \right)r}}{{\sqrt 3 }}$
Thus, the radius of the circle which touching by all the three given circles internally is $R = \dfrac{{\left( {\sqrt 3 + 2} \right)r}}{{\sqrt 3 }}$
Note:
The above question can be solved with another method given below:
$\Delta DEF$ is equilateral with side 2r
Let us assume that the radius of circumference of triangle DEF be ${R_1}$ .
Now,
Area of $\Delta DEF = \dfrac{{\sqrt 3 }}{4}{\left( {2r} \right)^2}$
$\therefore $ Area of $\Delta DEF = \sqrt 3 {r^2}$
$\sqrt 3 {r^2} = \dfrac{{2r.2r.2r}}{{4{R_1}}}$
$\therefore {R_1} = \dfrac{{2r}}{{\sqrt 3 }}$
$\therefore {R_1} = r\dfrac{{\left( {2 + \sqrt 3 } \right)}}{{\sqrt 3 }}$
Thus, the radius of the circle which touching by all the three given circles internally is $R = \dfrac{{\left( {\sqrt 3 + 2} \right)r}}{{\sqrt 3 }}$
