The length of the chord joining points $(4 \cos \theta, 4 \sin \theta)$ and $\left[4 \cos…
The length of the chord joining points $(4 \cos \theta, 4 \sin \theta)$ and $\left[4 \cos \left(\theta+60^{\circ}\right)\right.$, $\left.4 \sin \left(\theta+60^{\circ}\right)\right]$ on the circle $x^2+y^2=16$ is
4
8
16
2
Solution
Given, equation of circle $x^2+y^2=16$
Points are $(4 \cos \theta, 4 \sin \theta)$ and
$
\left[4 \cos \left(\theta+60^{\circ}\right), 4 \sin \left(\theta+60^{\circ}\right)\right]
$
Clearly $\triangle O A B$ is an equilateral triangle.
$
\begin{aligned}
& \therefore \quad A B=O A=O B=r \\
& \text { where } r=4 (radius of circle)\\
& \Rightarrow \quad A B=4
\end{aligned}
$
