The equation of the chord of the circle `x^(2)+y^(2)-4x+6y-3=0` having (1,-2) as it midpoint is

Options

A. `x+y+1=0`
B. `2x+3y+4=0`
C. `x-y-3=0`
D. not existing

Answer

To find the equation of the chord of the circle given by the equation \(x^2 + y^2 - 4x + 6y - 3 = 0\) with midpoint \((1, -2)\), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we can rewrite the given circle equation in standard form. The equation is:

\[ x^2 + y^2 - 4x + 6y - 3 = 0 \]

We can complete the square for the \(x\) and \(y\) terms.

### Step 2: Completing the Square
For \(x\):

\[ x^2 - 4x \quad \text{can be rewritten as} \quad (x - 2)^2 - 4 \]

For \(y\):

\[ y^2 + 6y \quad \text{can be rewritten as} \quad (y + 3)^2 - 9 \]

Substituting these back into the circle equation gives:

\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0 \]

This simplifies to:

\[ (x - 2)^2 + (y + 3)^2 - 16 = 0 \]

Thus, the equation of the circle is:

\[ (x - 2)^2 + (y + 3)^2 = 16 \]

This indicates that the center of the circle is at \((2, -3)\) and the radius is \(4\).

### Step 3: Use the Midpoint Formula
The midpoint of the chord is given as \((1, -2)\). We denote this point as \((x_1, y_1) = (1, -2)\).

### Step 4: Write the Equation of the Chord
The equation of the chord with midpoint \((x_1, y_1)\) can be found using the formula:

\[ S_1 = S(x_1, y_1) = 0 \]

Where \(S(x, y)\) is derived from the circle's equation.

### Step 5: Calculate \(S_1\)
Substituting \(x_1 = 1\) and \(y_1 = -2\) into the circle equation:

\[ S_1 = x \cdot 1 + y \cdot (-2) - 4 \cdot \frac{1}{2} + 6 \cdot \frac{-2}{2} - 3 = 0 \]

This simplifies to:

\[ x - 2y - 2 - 6 - 3 = 0 \]

Combining like terms gives:

\[ x - 2y - 11 = 0 \]

### Step 6: Calculate \(S_{11}\)
Next, we calculate \(S_{11}\) using the coordinates of the midpoint:

\[ S_{11} = (1)^2 + (-2)^2 - 4(1) + 6(-2) - 3 \]

Calculating this gives:

\[ 1 + 4 - 4 - 12 - 3 = -14 \]

### Step 7: Set \(S_1 = S_{11}\)
Now we equate \(S_1\) and \(S_{11}\):

\[ -14 = x - 2y - 11 \]

Rearranging gives:

\[ x - 2y + 11 - 14 = 0 \implies x - 2y - 3 = 0 \]

### Final Answer
Thus, the equation of the chord is:

\[ \boxed{x - 2y - 3 = 0} \]