Question

Given two circles intersecting orthogonally having the length of common chord $24 / 5$ units. The radius of one of the circles is 3 units.
The length of direct common tangent is.

• A. $\sqrt{12}$
• B. $4 \sqrt{3}$
• C. $2 \sqrt{6}$
• D. $3 \sqrt{6}$

Answer 1

Answer: (C)

We have $\sin \phi= d / r _{1}$ and $\cos \phi= d / r _{2}$, (where $2 d$ is the length of the common chord). Then,
$1=\frac{ d ^{2}}{ r _{1}^{2}}+\frac{ d ^{2}}{ r _{2}^{2}}$
or $d =\frac{ r _{1} r _{2}}{\sqrt{ r _{1}^{2}+ r _{2}^{2}}}$
or $2 d =\frac{2 r _{1} r _{2}}{\sqrt{ r _{1}^{2}+ r _{2}^{2}}}=\frac{24}{5}$, where $r _{1}=3$
or $d =\frac{6 r _{2}}{\sqrt{9+ r _{2}^{2}}}=\frac{24}{5}$
or $r_{2}=4$
From the gigure,
$\sin \frac{\theta}{2}=\frac{r_{2}-r_{1}}{C_{1} C_{2}}$
Where $C_{1} C_{2}^{2}=r_{1}^{2}+r_{1}^{2}$
or $C_{1} C_{2}=5$,
$\therefore \sin \frac{\theta}{2}=\frac{1}{5}$
or $\cos \frac{\theta}{2}=\frac{\sqrt{24}}{5}$
$\therefore \sin \theta=2 \times \frac{1}{5} \times \frac{\sqrt{24}}{5} $
$=\frac{2 \sqrt{6}}{25}$
or $\theta=\sin ^{-1} \frac{4 \sqrt{6}}{25}$
Also, $AB =\sqrt{ C _{1} C _{2}^{2}-\left( r _{1}- r _{2}\right)^{2}}$
$=\sqrt{25-1}=\sqrt{24}$