Given that, the length of long chord is $6 \; \text{cm},$ the length of small chord is $4 \; \text{cm},$ and the distance between these chords is $1\;\text{cm}.$
Let the radius of the circle be $ `r\text{’} \; \text{cm},$ and let the distance of long chord from the center of circle be $ `k\text{’} \; \text{cm}.$
We can draw the diagram,
We know that the perpendicular from the center of the circle to the chords bisects the chords.
In $\triangle \text{OFB},$ apply the Pythagoras’ theorem :
$ \text{(OB)}^{2} = \text{(OF)}^{2} + \text{(FB)}^{2}$
$ \Rightarrow r^{2} = (k+1)^{2} + 2^{2}$
$ \Rightarrow r^{2} = k^{2} + 1 + 2k + 4$
$ \Rightarrow r^{2} = k^{2} + 2k + 5 \quad \longrightarrow (1)$
In $ \triangle \text{OED},$ apply the Pythagoras’ theorem :
$ \text{(OD)}^{2} = \text{(OE)}^{2} + \text{(ED)}^{2}$
$ \Rightarrow r^{2} = k^{2} + 3^{3} $
$ \Rightarrow r^{2} = k^{2} + 9 \quad \longrightarrow (2)$
On equating the equation $(1),$ and $(2),$ we get,
$ k^{2} + 2k + 5 = k^{2} + 9 $
$ \Rightarrow 2k = 4 $
$ \Rightarrow \boxed {k= 2 \;\text{cm}}$
Put the value of $k$ in equation $(2),$ we get,
$ r^{2} = 2^{2} + 9$
$ \Rightarrow r^{2} = 4 + 9$
$ \Rightarrow r^{2} = 13$
$\Rightarrow \boxed { r = \sqrt{13}\; \text{cm}}$
$\therefore$ The radius of the circle is $\sqrt{13}\;\text{cm}.$
Correct Answer $: \text{C}$