The centre of circle whose normals are `x^(2)-2xy-3x+6y=0` is

Options

A. `(3,(3)/(2))`
B. `(3,-(3)/(2))`
C. `((3)/(2),3)`
D. None of these

Answer

To find the center of the circle whose normals are given by the equation \( x^2 - 2xy - 3x + 6y = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ x^2 - 2xy - 3x + 6y = 0 \]

### Step 2: Factoring the Equation
We can rearrange the equation by grouping terms:
\[ x^2 - 2xy - 3x + 6y = 0 \]
We can factor out \( x \) from the first two terms and \( -3 \) from the last two terms:
\[ x(x - 2y) - 3(x - 2y) = 0 \]
This gives us:
\[ (x - 2y)(x - 3) = 0 \]

### Step 3: Finding the Values of \( x \)
From the factored form, we can set each factor to zero:
1. \( x - 2y = 0 \) → \( x = 2y \)
2. \( x - 3 = 0 \) → \( x = 3 \)

### Step 4: Finding the Corresponding Values of \( y \)
Now, we substitute \( x = 3 \) into the equation \( x = 2y \) to find \( y \):
\[ 3 = 2y \implies y = \frac{3}{2} \]

### Step 5: Finding the Center of the Circle
Now we have the values of \( x \) and \( y \):
- \( x = 3 \)
- \( y = \frac{3}{2} \)

Thus, the center of the circle is:
\[ (3, \frac{3}{2}) \]

### Final Answer
The center of the circle is \( (3, \frac{3}{2}) \).

---