Two circles $C_1$ and $C_2$ of radius $2$ and $3$ respectively touch each other as shown in the figure .If $AD$ and $BD$ are tangents then the length of $BD$ is
$a.)3\sqrt6\\ b.)5\sqrt6\\ \color{green}{c.)\dfrac{7\sqrt6}{3}}\\ d.)6\\$
I did a construction of $CE$
And with help of Pythagorus found $AE=2\sqrt{10}$
And with again pythagorus i applied
$10^2+x^2=(x+2\sqrt{10})^2\implies x=3 \sqrt{\dfrac52}$
But the book is giving option $c.$ what is the mistake ?
On
As shown in the second diagram, we have in right $\Delta ABD$ we have $$AD=\sqrt{x^2+10^2}=\sqrt{x^2+100}$$ Now, in right triangles $\Delta ABD$ & $\Delta AEC_{2}$, we have $$\sin \angle DAB=\sin \angle EAC_{2} \implies \frac{BD}{AD}=\frac{EC_{2}}{AC_{2}} \implies \frac{x}{\sqrt{x^2+100}}=\frac{3}{7}$$ $$49x^2-9x^2=900 \implies x=\sqrt{\frac{900}{40}}=\sqrt{\frac{45}{2}}=3\sqrt{\frac{5}{2}}$$
The result is obviously same as you have obtained hence, there is certainly some printing mistake in the options provided in your book.
$$\sin\angle{EAC} = \frac{R}{2r+R} \implies \angle{EAC} = 25.37$$
$$\tan\angle{DAB} = \frac{x}{2r+2R} \implies x = 4.74$$
You are right, there is a mistake in the book.