`f(x , y)=x^2+y^2+2a x+2b y+c=0` represents a circle. If `f(x ,0)=0` has equal roots, each being `2,` and `f(0,y)=0` has 2 and 3 as its roots, then the center of the circle is

Options

A. `(2,5//2)`
B. Data are not sufficient
C. `(-2,-5//2)`
D. Data are inconsistent.

Explanation

To solve the problem, we need to analyze the given function \( f(x, y) = x^2 + y^2 + 2ax + 2by + c = 0 \) and use the conditions provided for the roots.

### Step 1: Analyze \( f(x, 0) = 0 \)

When \( y = 0 \), the function simplifies to:
\[ f(x, 0) = x^2 + 2ax + c = 0 \]
We are given that this has equal roots, each being \( 2 \). Therefore, we can express this quadratic equation as:
\[ (x - 2)^2 = 0 \]
Expanding this gives:
\[ x^2 - 4x + 4 = 0 \]
From this, we can identify that:
\[ 2a = 4 \quad \text{and} \quad c = 4 \]
Thus, we find:
\[ a = 2 \quad \text{and} \quad c = 4 \]

### Step 2: Analyze \( f(0, y) = 0 \)

Next, when \( x = 0 \), the function simplifies to:
\[ f(0, y) = y^2 + 2by + c = 0 \]
We are given that this has roots \( 2 \) and \( 3 \). Therefore, we can express this quadratic equation as:
\[ (y - 2)(y - 3) = 0 \]
Expanding this gives:
\[ y^2 - 5y + 6 = 0 \]
From this, we can identify that:
\[ 2b = -5 \quad \text{and} \quad c = 6 \]
Thus, we find:
\[ b = -\frac{5}{2} \quad \text{and} \quad c = 6 \]

### Step 3: Solve for \( c \)

Now, we have two different values for \( c \):
1. From \( f(x, 0) \), we found \( c = 4 \).
2. From \( f(0, y) \), we found \( c = 6 \).

This indicates an inconsistency in the data provided, as both conditions cannot hold simultaneously.

### Step 4: Find the center of the circle

The center of the circle represented by the equation \( f(x, y) = 0 \) can be found using the values of \( a \) and \( b \):
\[ \text{Center} = (-a, -b) = (-2, \frac{5}{2}) \]

### Conclusion

Thus, the center of the circle is:
\[ \text{Center} = (-2, \frac{5}{2}) \]