If two parallel chords of a circle, having diameter 4units, lie on the opposite sides of the center and subtend angles \cos ^ - 1( 1 \over 7 ) and sec - 1 (7) at the center respectivey, then the distance between these chords, is :

Circle PYQ — JEE Main 2017 · 8 Apr · Shift 1 · Q41 with Solution

View written solutionFree

Correct answer: B

Given diameter $=4$ , so radius is $r=\frac{4}{2}=2.$

If a chord subtends angle $\theta$ at the center, then its perpendicular distance from the center is $d=r\cos\left(\frac{\theta}{2}\right).$

Here the two chords subtend angles $\theta_1=\cos^{-1}\left(\frac{1}{7}\right), \qquad \theta_2=\sec^{-1}(7)=\cos^{-1}\left(\frac{1}{7}\right).$

So both angles are equal: $\theta_1=\theta_2=\cos^{-1}\left(\frac{1}{7}\right).$

Hence both chords are at the same perpendicular distance from the center: $d=2\cos\left(\frac{\theta}{2}\right).$

Using the half-angle formula, $\cos\left(\frac{\theta}{2}\right)=\sqrt{\frac{1+\cos\theta}{2}}.$

Since $\cos\theta=\frac{1}{7},$ we get

\sqrt{\frac{\frac{8}{7}}{2}}= \sqrt{\frac{4}{7}}=\frac{2}{\sqrt7}.$$ Therefore, $$d=2\cdot \frac{2}{\sqrt7}=\frac{4}{\sqrt7}.$$ 4. **Distance between the two parallel chords** The chords lie on opposite sides of the center, so the distance between them is the sum of their distances from the center: $$D=d+d=2d=2\cdot \frac{4}{\sqrt7}=\frac{8}{\sqrt7}.$$ 5. **Final answer** $$\boxed{\frac{8}{\sqrt7}}$$ So the correct option is **B**.