If a circle circle by assuming a chord parallel to the transverse axis of hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ as diameter always passes through $(2,0)$, then
Step-by-Step Solution
Key Concept: For a condition to hold for all values of a parameter, the coefficients of independent powers of that parameter must all equal zero.
Let the end points of the diameter of circle be $(a \sec \theta, b \tan \theta)$ and $(−a \sec \theta, b \tan \theta)$. The equation of circle is $(x − a \sec \theta)(x + a \sec \theta) + (y − b \tan \theta)^2 = 0$. Simplifying: $x^2 − a^2 \sec^2 \theta + y^2 + b^2 \tan^2 \theta − 2b \tan \theta y = 0$. Because equation (2.1) satisfies the above equation: $4 − a^2(1 + \tan^2 \theta) + 0 + b^2 \tan^2 \theta = 0 \Rightarrow (4 − a^2) + (b^2 − a^2)\tan^2 \theta = 0$. Which is always true if $|a| = 2 \& |b| = 2$.
Correct Answer: 2