Question

If $A \left(\frac{\pi}{3}\right), B \left(\frac{\pi}{6}\right) $ are the points on the circle represented in parametric form with centre $(0, 0)$ and radius $12$ then the length of the chord $AB$ is

• A. $6 ( \sqrt{6} - \sqrt{2} )$
• B. $6 ( \sqrt{6} - \sqrt{3} )$
• C. $\sqrt{2} ( \sqrt{3} - 2 )$
• D. $6 ( \sqrt{3} - 1)$

Answer 1

Answer: (A)

Parametric equations of given circle is $x=12 \cos \theta, y=12 \sin \theta$
$[\because.$ Parametric equation of $x^{2}+y^{2}=r^{2}$ is $x=r \cos \theta, y=r \sin \theta]$
Now, coordinates of point $A$ are given by
$x=12 \cos \frac{\pi}{3}, y=12 \sin \frac{\pi}{3}$
$\Rightarrow x=12 \cdot \frac{1}{2}, y=12 \cdot \frac{\sqrt{3}}{2}$
$\Rightarrow x=6 ; y=6 \sqrt{3}$
i.e. $A \equiv(6,6 \sqrt{3})$

and coordinates of point $B$ are given by
$x=12 \cos \frac{\pi}{6}, y=12 \sin \frac{\pi}{6}$
$\Rightarrow x=12 \cdot \frac{\sqrt{3}}{2} \cdot y=12 \cdot \frac{1}{2}$
$\Rightarrow x=6 \sqrt{3}, y=6$
i.e. $B \equiv(6 \sqrt{3}, 6)$
Clearly, length of chord
$A B=\sqrt{(6 \sqrt{3}-6)^{2}+(6-6 \sqrt{3})^{2}}$
$=\sqrt{2 \times 6^{2}(\sqrt{3}-1)^{2}}$
$=6 \sqrt{2}(\sqrt{3}-1)$
$=6(\sqrt{6}-\sqrt{2})$