If a circle drawn by assuming a chord parallel to the transverse axis of hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` as diameter always pases through (2, 0), then

Options

A. `|a|=|b|=2`
B. `|b| ne |a|`
C. `|b|=|a|=1`
D. `|b|=|a|=3`

Answer

To solve the problem, we need to analyze the given hyperbola and the conditions for the circle drawn using a chord parallel to the transverse axis. Let's break down the solution step by step.

### Step 1: Understand the Hyperbola
The equation of the hyperbola is given by:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

This hyperbola opens along the x-axis, and its transverse axis is horizontal.

### Step 2: Identify the Chord
We need to consider a chord of the hyperbola that is parallel to the transverse axis. Let the endpoints of this chord be \( A(-a \sec \theta, b \tan \theta) \) and \( B(a \sec \theta, b \tan \theta) \). The y-coordinates of points A and B are the same, which is \( b \tan \theta \).

### Step 3: Equation of the Circle
The circle with diameter AB can be expressed using the midpoint formula. The center of the circle is:

\[ \left(0, b \tan \theta\right) \]

The radius \( r \) of the circle is half the distance between points A and B, which is:

\[ r = \frac{1}{2} \left| a \sec \theta - (-a \sec \theta) \right| = a \sec \theta \]

The equation of the circle is:

\[ \left(x - 0\right)^2 + \left(y - b \tan \theta\right)^2 = (a \sec \theta)^2 \]

This simplifies to:

\[ x^2 + \left(y - b \tan \theta\right)^2 = a^2 \sec^2 \theta \]

### Step 4: Condition for the Circle to Pass through (2, 0)
For the circle to pass through the point (2, 0), we substitute \( x = 2 \) and \( y = 0 \) into the circle's equation:

\[ 2^2 + \left(0 - b \tan \theta\right)^2 = a^2 \sec^2 \theta \]

This simplifies to:

\[ 4 + b^2 \tan^2 \theta = a^2 \sec^2 \theta \]

### Step 5: Use the Identity
Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \), we can rewrite the equation:

\[ 4 + b^2 \tan^2 \theta = a^2 (1 + \tan^2 \theta) \]

Rearranging gives:

\[ 4 = a^2 - b^2 \tan^2 \theta + a^2 \tan^2 \theta \]

This can be factored as:

\[ 4 = a^2 - (b^2 - a^2) \tan^2 \theta \]

### Step 6: Analyze the Conditions
For the equation to hold for all values of \( \tan \theta \), the coefficient of \( \tan^2 \theta \) must be zero, leading to:

\[ b^2 - a^2 = 0 \implies b^2 = a^2 \implies |b| = |a| \]

Additionally, we also need to satisfy the condition when \( \tan^2 \theta \) is not involved, which gives:

\[ a^2 = 4 \implies |a| = 2 \]

### Conclusion
Thus, we conclude that:

\[ |a| = |b| = 2 \]