Question

Circle C₁ has a radius of 3 units. The line segment PQ is the only diameter of the circle which is parallel to the X-axis. P and Q are points on curves given by the equations y = a^x and y = 2a^x respectively, where a < 1. The value of a is:

• A. $\dfrac{1}{\sqrt[6]{2}}$
• B. $\dfrac{1}{\sqrt[6]{3}}$
• C. $\dfrac{1}{\sqrt[3]{6}}$
• D. $\dfrac{1}{\sqrt{6}}$
• E. None of the above

Hint:

When two curves have the same $y$-value, convert to logarithms to measure horizontal distance. For $y=a^{x}$ and $y=2a^{x}$, the horizontal gap is $\log_a 2$, independent of $y$—perfect for matching a given horizontal length.

Answer 1

The Correct Option is A

Step 1: Use that PQ is horizontal.
Since PQ is a diameter parallel to the X-axis, the y-coordinates of P and Q are equal.
Let this common ordinate be y > 0. Then:
P: (x₁, y) with y = a^x₁,   Q: (x₂, y) with y = 2a^x₂ ⇒ a^x₂ = y/2.

Step 2: Horizontal separation between the curves.
x₁ = log_a(y),    x₂ = log_a(y/2).
Hence the horizontal distance:
|x₁ − x₂| = |log_a(y) − log_a(y/2)| = |log_a(2)|.
This gap is constant, independent of y.

Step 3: Use the diameter length.
Radius of the circle is 3 ⇒ diameter PQ = 6. Since PQ is horizontal,
|log_a 2| = 6.
With a < 1, log_a 2 < 0, so log_a 2 = −6.
⇒ (ln 2)/(ln a) = −6 ⇒ ln a = −(1/6) ln 2 ⇒ a = 2^−1/6 = 1/√[6]{2}.

Final Answer:
a = 1 / √[6]{2}