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
$AB=\sqrt{(6\sqrt{3}-6{)}^2+(6-6\sqrt{3}{)}^2}$
$=\sqrt{2\times 6^2(\sqrt{3}-1{)}^2}$
$=6\sqrt{2}(\sqrt{3}-1)$
$=6(\sqrt{6}-\sqrt{2})$