Question

The locus of the mid points of the chords of the circle $C_1:(x-4)^2+(y-5)^2=4$ which subtend an angle $\theta_i$ at the centre of the circle $C_1$, is a circle of radius $r_i$ If $\theta_1=\frac{\pi}{3}, \theta_3=\frac{2 \pi}{3}$ and $r_1^2=r_2^2+r_3^2$, then $\theta_2$ is equal to

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{6}$
• D. $\frac{3 \pi}{4}$

Answer 1

The Correct Option is A

The correct answer is (A) : $\frac{\pi}{2}$

In $△CPB$

$\cos \frac{\theta }{2}=\frac{PC}{2}\Rightarrow PC=2\cos \frac{\theta }{2}$
$\Rightarrow (h-4{)}^2+(k-5{)}^2=4{\cos }^2\frac{\theta }{2}$
Now $(x-4{)}^2+(y-5{)}^2={\left(2\cos \frac{\theta }{2}\right)}^2$
$\Rightarrow r_1=2\cos \frac{\pi }{6}=\sqrt{3}$
$r_2=2\cos \frac{{\theta }_2}{2}$
$r_3=2\cos \frac{\pi }{3}=1$
$\Rightarrow r_1^2=r_2^2+r_3^2$
$\Rightarrow 3=4{\cos }^2\frac{{\theta }_2}{2}+1$
$\Rightarrow 4{\cos }^2\frac{{\theta }_2}{2}=2$
$\Rightarrow {\cos }^2\frac{{\theta }_2}{2}=\frac{1}{2}$
$\Rightarrow {\theta }_2=\frac{\pi }{2}$

Answer 2

Step 1: Understand the given geometry

The given circle \( C_1 : (x - 4)^2 + (y - 5)^2 = 4 \) has a radius of 2. 

The locus of the midpoints of the chords subtending an angle \( \theta_1 \) at the center is another circle with radius:

\[ r_1 = 2 \sin \left( \frac{\theta_1}{2} \right). \]

Similarly, the radii \( r_2 \) and \( r_3 \) correspond to angles \( \theta_2 \) and \( \theta_3 \), respectively.

Step 2: Relate \( r_1 \), \( r_2 \), and \( r_3 \)

The problem states:

\[ r_1^2 = r_2^2 + r_3^2. \] Using the formula for the radius of the locus, we have: \[ r_1 = 2 \sin \left( \frac{\theta_1}{2} \right), \quad r_2 = 2 \sin \left( \frac{\theta_2}{2} \right), \quad r_3 = 2 \sin \left( \frac{\theta_3}{2} \right). \] Substitute these into the equation: \[ 2 \sin \left( \frac{\theta_1}{2} \right)^2 = 2 \sin \left( \frac{\theta_2}{2} \right)^2 + 2 \sin \left( \frac{\theta_3}{2} \right)^2. \] Simplify: \[ 4 \sin^2 \left( \frac{\theta_1}{2} \right) = 4 \sin^2 \left( \frac{\theta_2}{2} \right) + 4 \sin^2 \left( \frac{\theta_3}{2} \right). \] Divide through by 4: \[ \sin^2 \left( \frac{\theta_1}{2} \right) = \sin^2 \left( \frac{\theta_2}{2} \right) + \sin^2 \left( \frac{\theta_3}{2} \right). \] 

Step 3: Substitute known values of \( \theta_1 \) and \( \theta_3 \)

From the problem, \( \theta_1 = \frac{\pi}{3} \) and \( \theta_3 = \frac{2\pi}{3} \).

Calculate \( \sin^2 \left( \frac{\theta_1}{2} \right) \):

\[ \sin^2 \left( \frac{\pi}{6} \right) = \left( \frac{1}{2} \right)^2 = \frac{1}{4}. \]

Calculate \( \sin^2 \left( \frac{\theta_3}{2} \right) \):

\[ \sin^2 \left( \frac{\pi}{3} \right) = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4}. \]

Substitute these into the equation:

\[ \frac{1}{4} = \sin^2 \left( \frac{\theta_2}{2} \right) + \frac{3}{4}. \] Solve for \( \sin^2 \left( \frac{\theta_2}{2} \right) \): \[ \sin^2 \left( \frac{\theta_2}{2} \right) = \frac{1}{4} - \frac{3}{4} = \frac{1}{2}. \] 

Step 4: Determine \( \theta_2 \)

If \( \sin^2 \left( \frac{\theta_2}{2} \right) = \frac{1}{2} \), then:

\[ \sin \left( \frac{\theta_2}{2} \right) = \frac{\sqrt{2}}{2}. \] This corresponds to: \[ \frac{\theta_2}{2} = \frac{\pi}{4} \quad \Rightarrow \quad \theta_2 = \frac{\pi}{2}. \]