JEE Mathematics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: The circle is $C_1: (x - 4)^2 + (y - 5)^2 = 4$, so its radius is $2$. Also, $\theta_1 = \frac{\pi}{3}$, $\theta_3 = \frac{2\pi}{3}$, and $r_1^2 = r_2^2 + r_3^2$.

Find: $\theta_2$.

For a chord subtending angle $\theta$ at the centre, the midpoint of the chord lies at distance

$$PC = 2\cos\left(\frac{\theta}{2}\right)$$

from the centre. Hence the locus of such midpoints is a circle centered at $(4,5)$ with radius

$$r = 2\cos\left(\frac{\theta}{2}\right).$$

Therefore,

$$r_1 = 2\cos\left(\frac{\theta_1}{2}\right), \quad r_2 = 2\cos\left(\frac{\theta_2}{2}\right), \quad r_3 = 2\cos\left(\frac{\theta_3}{2}\right).$$

Using $\theta_1 = \frac{\pi}{3}$,

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

Using $\theta_3 = \frac{2\pi}{3}$,

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

Now apply

$$r_1^2 = r_2^2 + r_3^2.$$

So,

$$3 = r_2^2 + 1$$

which gives

$$r_2^2 = 2.$$

Since

$$r_2 = 2\cos\left(\frac{\theta_2}{2}\right),$$

we get

$$4\cos^2\left(\frac{\theta_2}{2}\right) = 2.$$

Hence,

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

Therefore,

$$\frac{\theta_2}{2} = \frac{\pi}{4}$$

and so

$$\theta_2 = \frac{\pi}{2}.$$

The correct option is D.

Common mistakes

• Using $r = 2\sin\left(\frac{\theta}{2}\right)$ for the locus radius. That gives the half-chord length, not the distance of the midpoint from the centre. Here the required radius is $PC = 2\cos\left(\frac{\theta}{2}\right)$.

• Forgetting that the given circle has radius $2$, not $1$. The formula must be multiplied by the actual circle radius, so $r = 2\cos\left(\frac{\theta}{2}\right)$.

• Substituting $\theta$ instead of $\frac{\theta}{2}$ in the trigonometric expression. The right triangle at the midpoint uses half of the central angle, so the argument must be $\frac{\theta}{2}$.