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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 $: \text{C}$