Question

If the circles \(x^2+y^2 = 16\) and \(x^2+y^2+2ax+4y+4 = 0\) touch each other internally then \(a =\)

• A. \(\frac{2}{3}\)
• B. \(\frac{1}{56}\)
• C. \(\frac{-2}{3}\)
• D. \(\frac{3}{2}\)

Hint:

Multiply and divide by sin theta and use the double angle formula repeatedly.

Answer 1

The Correct Option is D

Step 1: Key Formula or Approach:
\[ \cos\theta\cos 2\theta\cos 4\theta\cdots\cos 2^{n-1}\theta = \frac{\sin 2^n\theta}{2^n\sin\theta} \]
This comes from repeated use of \(2\sin A\cos A = \sin 2A\).

Step 2: Apply for five factors:
Here the factors go up to \(\cos 16\theta\), so \(n = 5\):
\[ \cos\theta\cos2\theta\cos4\theta\cos8\theta\cos16\theta = \frac{\sin 32\theta}{32\sin\theta} \]

Step 3: Use the given condition:
Since \(33\theta = \pi\), \(32\theta = \pi - \theta\) and \(\sin 32\theta = \sin(\pi - \theta) = \sin\theta\).
\[ \text{Product} = \frac{\sin\theta}{32\sin\theta} = \frac{1}{32} \]
The sign is positive since \(\sin\theta \ne 0\) and the identity holds without any sign change.

Final Answer:
The value is \(\frac{1}{32}\), option (B). \[ \boxed{\frac{1}{32}} \]