Key Concepts and Formulas

• The locus of the midpoints of chords of a circle that subtend a constant angle at the center is a concentric circle.
• Relationship between the radius of the locus circle ($r_i$), the radius of the original circle ($R$), and the angle subtended at the center ($\theta_i$): $r_i = R\cos(\frac{\theta_i}{2})$.
• Trigonometric identities and values for common angles.

Step-by-Step Solution

Step 1: Understanding the Geometry and Deriving the Radius of the Locus

• What we are doing: We are given a circle and we want to find the locus of the midpoints of chords that subtend a specific angle at the center. We will use geometry and trigonometry to relate the radius of the circle formed by these midpoints to the angle subtended by the chords.
• Why: This is a standard problem and understanding how the radius of the smaller circle formed by the midpoints of the chords is related to the angle subtended at the center of the original circle is important.
• Math: The equation of the given circle $C_1$ is $(x - 4)^2 + (y - 5)^2 = 4$. This tells us that the center of the circle is $C(4, 5)$ and the radius is $R = \sqrt{4} = 2$.

Let $AB$ be a chord of the circle $C_1$ and let $M$ be the midpoint of $AB$. Then $CM$ is perpendicular to $AB$, and $\angle ACB = \theta_i$. Also, $\angle ACM = \frac{\theta_i}{2}$.

In right triangle $\triangle CMA$, we have $\cos(\angle ACM) = \frac{CM}{CA}$. Thus, $\cos(\frac{\theta_i}{2}) = \frac{CM}{R}$. Therefore, $CM = R \cos(\frac{\theta_i}{2})$.

Since the locus of $M$ is a circle with center $C$ and radius $r_i = CM$, we have $r_i = R \cos(\frac{\theta_i}{2})$. Since $R = 2$, we have $r_i = 2 \cos(\frac{\theta_i}{2})$.

Step 2: Applying the Given Conditions and Relationship

• What we are doing: Now that we have a formula for $r_i$ in terms of $\theta_i$, we will use the given values of $\theta_1$ and $\theta_3$ to find $r_1^2$ and $r_3^2$.
• Why: The problem gives us a relationship between $r_1^2$, $r_2^2$, and $r_3^2$. By finding $r_1^2$ and $r_3^2$, we can solve for $r_2^2$.
• Math: We are given $\theta_1 = \frac{\pi}{3}$. Then $r_1 = 2 \cos(\frac{\pi/3}{2}) = 2 \cos(\frac{\pi}{6}) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$. Therefore, $r_1^2 = 3$.

We are given $\theta_3 = \frac{2\pi}{3}$. Then $r_3 = 2 \cos(\frac{2\pi/3}{2}) = 2 \cos(\frac{\pi}{3}) = 2 \cdot \frac{1}{2} = 1$. Therefore, $r_3^2 = 1$.

We are given that $r_1^2 = r_2^2 + r_3^2$. Substituting the values we found, we get $3 = r_2^2 + 1$. Thus, $r_2^2 = 2$.

Step 3: Solving for $\theta_2$

• What we are doing: We now know $r_2^2$, and we have a formula relating $r_2$ to $\theta_2$. We can solve for $\theta_2$.
• Why: This will give us the final answer, the angle subtended by the chords corresponding to the locus circle with radius $r_2$.
• Math: We have $r_2 = 2 \cos(\frac{\theta_2}{2})$. Squaring both sides, we get $r_2^2 = 4 \cos^2(\frac{\theta_2}{2})$.

Since $r_2^2 = 2$, we have $2 = 4 \cos^2(\frac{\theta_2}{2})$. Then $\cos^2(\frac{\theta_2}{2}) = \frac{1}{2}$. Taking the square root, we get $\cos(\frac{\theta_2}{2}) = \pm \frac{1}{\sqrt{2}}$.

Since $0 < \theta_i < \pi$, we have $0 < \frac{\theta_i}{2} < \frac{\pi}{2}$. Therefore, $\cos(\frac{\theta_i}{2})$ must be positive. So $\cos(\frac{\theta_2}{2}) = \frac{1}{\sqrt{2}}$.

Thus, $\frac{\theta_2}{2} = \frac{\pi}{4}$, which means $\theta_2 = \frac{\pi}{2}$.

Common Mistakes & Tips

• Remember that the angle $\theta_i$ must be in radians.
• Be careful with trigonometric identities and values.
• Always remember the range of the inverse cosine function when finding angles.

Summary

We were given the locus of midpoints of chords of a circle which subtend an angle ${\theta _i}$ at the centre of the circle $C_1$, is a circle of radius $r_i$. After deriving the formula of radius $r_i$ as a function of $\theta_i$, we used the relationship between the radii to solve for $\theta_2$.

The final answer is $\frac{\pi}{2}$.

Final Answer

The final answer is \boxed{\frac{\pi}{2}}, which corresponds to option (A).