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

To solve the problem, we will first derive the general formula for the locus of the midpoints of chords of a circle that subtend a certain angle at the center. We will then use this information to determine the unknown angle, \( \theta_2 \).

The given circle is \( C_1: (x-4)^2 + (y-5)^2 = 4 \). The center of this circle is \( (4, 5) \) and its radius is 2.

It is a known property that the locus of midpoints of chords of a circle which subtend a constant angle \( \theta \) at the center is another circle, called the circle of Apollonius. The radius \( r_i \) of this circle can be given by:

\(r_i = R \cos\left(\frac{\theta_i}{2}\right)\)

where \( R \) is the radius of the original circle. In this case, \( R = 2 \).

Using the given data:

• For \( \theta_1 = \frac{\pi}{3} \),

\(r_1 = 2 \cos\left(\frac{\pi}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}\)

• For \( \theta_2 = \theta_2 \), we have: \( r_2 = 2 \cos\left(\frac{\theta_2}{2}\right)\)
• For \( \theta_3 = \frac{2\pi}{3} \),

\(r_3 = 2 \cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1\)

According to the problem, \( r_1^2 = r_2^2 + r_3^2 \).

\((\sqrt{3})^2 = (r_2)^2 + (1)^2\)

\(3 = (r_2)^2 + 1\)

\((r_2)^2 = 2\)

\(r_2 = \sqrt{2}\)

We know that:

\(\sqrt{2} = 2 \cos\left(\frac{\theta_2}{2}\right)\)

\(\cos\left(\frac{\theta_2}{2}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\)

This corresponds to \( \frac{\theta_2}{2} = \frac{\pi}{4} \), hence:

\(\theta_2 = \frac{\pi}{2}\)

The correct answer is \( \frac{\pi}{2} \), therefore, the option is \( \frac{\pi}{2} \).