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 $\sin \varphi =d/r_1$ and $\cos \varphi =d/r_2$, (where $2d$ 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 $2d=\frac{2r_1{r}_2}{\sqrt{r_1^2+r_2^2}}=\frac{24}{5}$, where $r_1=3$
or $d=\frac{6r_2}{\sqrt{9+r_2^2}}=\frac{24}{5}$
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, $AB=\sqrt{C_1{C}_2^2-{\left(r_1-r_2\right)}^2}$
$=\sqrt{25-1}=\sqrt{24}$