Locus of intersection of two perpendicular tangents to the hyperbola is:
$(x-3)^2 + \left(y - \frac{7}{2}\right)^2 = \frac{55}{4}$
$(x-3)^2 + \left(y - \frac{7}{2}\right)^2 = \frac{25}{4}$
$(x-3)^2 + \left(y - \frac{7}{2}\right)^2 = \frac{7}{4}$
None of these
Step-by-Step Solution
Key Concept: The locus of intersection of two perpendicular tangents to a hyperbola is its director circle, given by $(x-h)^2 + (y-k)^2 = a^2 - b^2$ for hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$. For a hyperbola, this exists only when $a^2 > b^2$, and if the condition fails, no real director circle exists.
The director circle of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is found using the centre (midpoint of foci) at $(h,k) = (3, \frac{7}{2})$ and the relation $b^2 = a^2(e^2-1)$. With $a = \frac{3}{2}$, $e = \frac{5}{3}$, we get $b^2 = 4$. The director circle equation becomes $(x-3)^2 + (y-\frac{7}{2})^2 = \frac{7}{4}$, which does not represent any real point.
Correct Answer: 4