Explanation

(b)
Given equation of line is
$3x - 2y = k$ …(i)
And equation of circle is
$x^{2} + y^{2} = 4r^{2}$ …(ii)
Eq. (i) can be rewritten as $y = \frac{3}{2}x - \frac{k}{2}$
$\Rightarrow m = \frac{3}{2},c = - \frac{k}{2}$
The line will meet the circle in one point, if
$c = a\sqrt{1 + m^{2}}$
$\Rightarrow \ - \frac{k}{2} = (2r)\sqrt{1 + \left( \frac{3}{2} \right)^{2}}$
On squaring, we get
$\frac{k^{2}}{4} = 4r^{2} \times \frac{13}{4}$
$\Rightarrow \ k^{2} = 52\ r^{2}$