Explanation

(c)
The equation of tangent at point (1, 2) to the circle $x^{2} + y^{2} - 4x - 6y + 9 = 0$, is
$x + 2y - 2(x + 1) - 3(y + 2) + 9 = 0$
$\Rightarrow \ \ \ x + y - 1 = 0$
Since, the inverse of the point (1, 2) is the foot $(\alpha,\ \beta)$ of the perpendicular from the point (1, 2) to the line $x + y - 1$
$\therefore\ \ \ \ \frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = - \frac{(1.1 + 1.2 - 1)}{1^{2} + 1^{2}}$
$\Rightarrow \ \ \ \alpha - 1 = \beta - 2 = - 1$
$\Rightarrow \ \alpha = 0,\ \beta = 1$
Hence, required point is (0, 1)