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 that is formed by 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 as:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
This hyperbola opens along the x-axis, meaning its transverse axis is horizontal.

### Step 2: Define the Chord
We need to consider a chord of the hyperbola that is parallel to the transverse axis (x-axis). Let's denote two points on the hyperbola as:
\[ P(a \sec \theta, b \tan \theta) \quad \text{and} \quad Q(-a \sec \theta, b \tan \theta) \]
These points represent the endpoints of the chord.

### Step 3: Find the Circle's Equation
The circle having chord \(PQ\) as its diameter can be expressed using the midpoint formula. The midpoint \(M\) of \(PQ\) is:
\[ M\left(0, b \tan \theta\right) \]
The radius of the circle is half the distance between points \(P\) and \(Q\):
\[ \text{Radius} = \frac{1}{2} \sqrt{(a \sec \theta - (-a \sec \theta))^2 + (b \tan \theta - b \tan \theta)^2} = a \sec \theta \]
Thus, the equation of the circle is:
\[ (x - 0)^2 + \left(y - b \tan \theta\right)^2 = (a \sec \theta)^2 \]
This simplifies to:
\[ x^2 + (y - b \tan \theta)^2 = a^2 \sec^2 \theta \]

### Step 4: Substitute the Point (2, 0)
We know that the circle passes through the point (2, 0). Substituting \(x = 2\) and \(y = 0\) into the circle's equation gives:
\[ 2^2 + (0 - b \tan \theta)^2 = a^2 \sec^2 \theta \]
This simplifies to:
\[ 4 + b^2 \tan^2 \theta = a^2 \sec^2 \theta \]

### Step 5: Use Trigonometric Identity
Using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), we can rewrite the equation:
\[ 4 + b^2 \tan^2 \theta = a^2 (1 + \tan^2 \theta) \]
This can be rearranged to:
\[ 4 = a^2 - b^2 \tan^2 \theta + a^2 \tan^2 \theta \]
\[ 4 = a^2 - (b^2 - a^2) \tan^2 \theta \]

### Step 6: Set Conditions for the Equation
For the equation to hold for all values of \(\theta\), the coefficients of \(\tan^2 \theta\) must equal zero:
1. \(b^2 - a^2 = 0 \Rightarrow b = a\)
2. \(4 = a^2\) or \(a = \pm 2\)

### Conclusion
Thus, we have two conditions:
1. \(a = b\)
2. \(|a| = 2\) and \(|b| = 2\)

### Final Answer
The relationship that satisfies the conditions is:
\[ |a| = |b| = 2 \]