Statement-1: If limiting points of a family of co-axial system of circles are (1, 1) and (3, 3), then `2x^(2)+2y^(2)-3x-3y=0` is a member of this family passing through the origin.
Statement-2: Limiting points of a family of coaxial circles are the centres of the circles with zero radius.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze both statements and verify their validity step by step.

### Step 1: Understanding Statement 2
**Statement 2** states that the limiting points of a family of coaxial circles are the centers of the circles with zero radius.

- **Explanation**: The limiting points of a family of coaxial circles are indeed the centers of those circles when the radius approaches zero. This is a fundamental property of coaxial circles.

**Conclusion**: Statement 2 is **true**.

### Step 2: Understanding Statement 1
**Statement 1** states that if the limiting points of a family of coaxial circles are (1, 1) and (3, 3), then the equation \(2x^2 + 2y^2 - 3x - 3y = 0\) is a member of this family passing through the origin.

- **Limiting Points**: The limiting points (1, 1) and (3, 3) are the centers of the circles in the family.

### Step 3: Writing the General Equation of Coaxial Circles
The general equation of coaxial circles can be expressed as a linear combination of the equations of the circles centered at the limiting points:

1. Circle centered at (1, 1):
\[ (x - 1)^2 + (y - 1)^2 = r^2 \]
When \(r = 0\), this becomes:
\[ (x - 1)^2 + (y - 1)^2 = 0 \]

2. Circle centered at (3, 3):
\[ (x - 3)^2 + (y - 3)^2 = r^2 \]
When \(r = 0\), this becomes:
\[ (x - 3)^2 + (y - 3)^2 = 0 \]

### Step 4: Forming the Family of Circles
The family of coaxial circles can be expressed as:
\[ (x - 1)^2 + (y - 1)^2 + \lambda \left[(x - 3)^2 + (y - 3)^2\right] = 0 \]
where \(\lambda\) is a parameter.

### Step 5: Finding the Value of \(\lambda\)
To find \(\lambda\), we need to check if the family passes through the origin (0, 0):
\[ (0 - 1)^2 + (0 - 1)^2 + \lambda \left[(0 - 3)^2 + (0 - 3)^2\right] = 0 \]
Calculating:
\[ 1 + 1 + \lambda(9 + 9) = 0 \]
This simplifies to:
\[ 2 + 18\lambda = 0 \]
Solving for \(\lambda\):
\[ 18\lambda = -2 \implies \lambda = -\frac{1}{9} \]

### Step 6: Substituting \(\lambda\) Back
Now we substitute \(\lambda\) back into the family equation:
\[ (x - 1)^2 + (y - 1)^2 - \frac{1}{9} \left[(x - 3)^2 + (y - 3)^2\right] = 0 \]

### Step 7: Simplifying the Equation
Expanding and simplifying:
1. Expand \((x - 1)^2 + (y - 1)^2\):
\[ x^2 - 2x + 1 + y^2 - 2y + 1 = x^2 + y^2 - 2x - 2y + 2 \]

2. Expand \((x - 3)^2 + (y - 3)^2\):
\[ x^2 - 6x + 9 + y^2 - 6y + 9 = x^2 + y^2 - 6x - 6y + 18 \]

3. Substitute into the equation:
\[ x^2 + y^2 - 2x - 2y + 2 - \frac{1}{9}(x^2 + y^2 - 6x - 6y + 18) = 0 \]
Multiply through by 9 to eliminate the fraction:
\[ 9(x^2 + y^2 - 2x - 2y + 2) - (x^2 + y^2 - 6x - 6y + 18) = 0 \]
Simplifying this leads to:
\[ 2x^2 + 2y^2 - 3x - 3y = 0 \]

### Conclusion
Since we derived the equation \(2x^2 + 2y^2 - 3x - 3y = 0\) from the family of coaxial circles, **Statement 1 is true**.

### Final Answer
Both statements are true, and Statement 2 is a correct explanation for Statement 1.

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