STATEMENT-1 : If n circles `(n ge 3)`, no two circles are non-centric and no three centre are collinear and number of radical centre is equal to number of radical axes, then n = 5.
and
STATEMENT-2 : If no three centres are collinear and no two circles are concentric, then number of radical centre is `""^(n)C_(3)` and number of radical axs is `""^(n)C_(2).`

Options

A. (a) Statement-1 is true, statement-2 is true, Statement -2 is a correct explanation for Statement -1
B. (b)Statement -1 is true, Statement-2 is true , Statement-2 is NOT a correct explanation for statement-1
C. (c)Statement-1 is true, Statement-2 is False
D. (d)Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we will analyze both statements step by step.

### Step 1: Understanding the Statements
- **Statement 1** asserts that for \( n \) circles (where \( n \geq 3 \)), if no two circles are concentric and no three centers are collinear, the number of radical centers equals the number of radical axes, then \( n = 5 \).
- **Statement 2** states that if no three centers are collinear and no two circles are concentric, then the number of radical centers is \( \binom{n}{3} \) and the number of radical axes is \( \binom{n}{2} \).

### Step 2: Analyzing Statement 2
- The number of radical centers formed by \( n \) circles is given by \( \binom{n}{3} \) because a radical center is formed by the intersection of the radical axes of any three circles.
- The number of radical axes formed by \( n \) circles is given by \( \binom{n}{2} \) since a radical axis is determined by any two circles.

### Step 3: Setting Up the Equation
From Statement 1, we have:
\[ \text{Number of radical centers} = \text{Number of radical axes} \]
This can be expressed as:
\[ \binom{n}{3} = \binom{n}{2} \]

### Step 4: Expanding the Binomial Coefficients
Using the formula for binomial coefficients:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
We can write:
\[ \binom{n}{3} = \frac{n!}{3!(n-3)!} \quad \text{and} \quad \binom{n}{2} = \frac{n!}{2!(n-2)!} \]

### Step 5: Setting Up the Equality
Setting the two expressions equal gives:
\[ \frac{n!}{3!(n-3)!} = \frac{n!}{2!(n-2)!} \]

### Step 6: Simplifying the Equation
We can cancel \( n! \) from both sides (assuming \( n \neq 0 \)):
\[ \frac{1}{3!(n-3)!} = \frac{1}{2!(n-2)!} \]
This simplifies to:
\[ \frac{1}{6(n-3)!} = \frac{1}{2(n-2)(n-3)!} \]
Multiplying both sides by \( 6(n-3)! \) gives:
\[ 1 = \frac{6}{2(n-2)} \]
This simplifies to:
\[ 1 = \frac{3}{n-2} \]

### Step 7: Solving for \( n \)
Cross-multiplying gives:
\[ n - 2 = 3 \implies n = 5 \]

### Conclusion
Thus, we conclude that Statement 1 is true and \( n = 5 \).

### Step 8: Verifying Statement 2
From our analysis, Statement 2 is also correct as it provides the correct expressions for the number of radical centers and radical axes.

### Final Answer
Both Statement 1 and Statement 2 are true.

---