## 1 Answer

Given that, the diameter of $\text{C1}$ is $2 \; \text{cm}$ longer than $\text{C2}.$

So, the radius of $\text{C1}$ should be $1 \; \text{cm}$ longer than $\text{C2}.$

Let the radius of $\text{C1} = \text{R}\; \text{cm}$

So, the radius of $\text{C2} = (\text{R} -1) \; \text{cm}$

The chord of $\text{C1}$ has length $6 \; \text{cm}$ and is a tangent to $\text{C2}.$

Now, we can draw the diagram,

In $\triangle \text{OCB},$ using the Pythagoras’ theorem,

$\text{(OB)}^{2} = \text{(OC)}^{2} + \text{(CB)}^{2} $

$ \Rightarrow \text{R}^{2} = \text{(R}-1)^{2} + 3^{2} $

$ \Rightarrow \text{R}^{2} = \text{R}^{2} + 1 – 2 \text{R} + 9 $

$ \Rightarrow 2 \text{R} = 10 $

$ \Rightarrow \boxed { \text{R} = 5 \; \text{cm}} $

So, diameter of $\text{C1} = 2 \text{R} = 2(5) = 10 \; \text{cm}.$

$\therefore$ The diameter of $\text{C1}$ is $10 \; \text{cm}.$

Correct Answer$: 10$