The equation of the chord of the circle `x^(2)+y^(2)-6x+8y=0` which is bisected at the point (5, -3), is

Options

A. 2x+y-7=0
B. x+2y+1=0
C. 2x-y-13=0
D. x-2y-11=0

Explanation

To find the equation of the chord of the circle \(x^2 + y^2 - 6x + 8y = 0\) which is bisected at the point (5, -3), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 6x + 8y = 0 \]
We can rearrange it as:
\[ x^2 - 6x + y^2 + 8y = 0 \]

### Step 2: Complete the Square
Next, we complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

For \(y\):
\[ y^2 + 8y = (y + 4)^2 - 16 \]

Now substituting back, we have:
\[ (x - 3)^2 - 9 + (y + 4)^2 - 16 = 0 \]
\[ (x - 3)^2 + (y + 4)^2 - 25 = 0 \]
\[ (x - 3)^2 + (y + 4)^2 = 25 \]
This represents a circle with center (3, -4) and radius 5.

### Step 3: Use the Midpoint Formula for Chords
The equation of a chord of a circle that is bisected at a point \((x_1, y_1)\) can be expressed as:
\[ T = S_1 \]
where \(T\) is the equation formed using the midpoint and \(S_1\) is the value of the circle's equation at the midpoint.

### Step 4: Identify \(T\) and \(S_1\)
Here, the midpoint is given as \((5, -3)\).

The general form of \(T\) is:
\[ T: x_1 x + y_1 y + g x + f y + c = 0 \]
From the circle's equation, we have:
- \(g = -3\) (from \(-6x\))
- \(f = 4\) (from \(8y\))
- \(c = 0\)

Substituting \(x_1 = 5\) and \(y_1 = -3\):
\[ T: 5x - 3y - 3x + 4y = 0 \]
This simplifies to:
\[ (5 - 3)x + (-3 + 4)y = 0 \]
\[ 2x + y = 0 \]

### Step 5: Calculate \(S_1\)
Now, we need to calculate \(S_1\) by substituting \((5, -3)\) into the circle's equation:
\[ S_1 = (5)^2 + (-3)^2 - 6(5) + 8(-3) \]
Calculating this:
\[ = 25 + 9 - 30 - 24 = -20 \]

### Step 6: Set \(T = S_1\)
Now we set \(T = S_1\):
\[ 2x + y = -20 \]

### Step 7: Rearranging the Equation
Rearranging gives us the final equation of the chord:
\[ 2x + y + 20 = 0 \]

### Final Answer
Thus, the equation of the chord of the circle which is bisected at the point (5, -3) is:
\[ \boxed{2x + y + 20 = 0} \]