Equation of a common chord of the circles `x ^(2) + y ^(2) + 6x -10 y + 9=0 and x ^(2) + y ^(2) - 10 x + 6y + 25=0 ` is

Options

A. `x + y + 4=0`
B. `x-y+ 4=0`
C. `x + y + 1=0`
D. `x - y -1=0`

Answer

To find the equation of the common chord of the circles given by the equations:

1. \( x^2 + y^2 + 6x - 10y + 9 = 0 \) (Circle 1)
2. \( x^2 + y^2 - 10x + 6y + 25 = 0 \) (Circle 2)

we can use the method of subtracting the equations of the two circles.

### Step 1: Write down the equations of the circles
The equations of the circles are:
- Circle 1: \( x^2 + y^2 + 6x - 10y + 9 = 0 \)
- Circle 2: \( x^2 + y^2 - 10x + 6y + 25 = 0 \)

### Step 2: Subtract the second equation from the first
We subtract the second equation from the first:
\[ (x^2 + y^2 + 6x - 10y + 9) - (x^2 + y^2 - 10x + 6y + 25) = 0 \]

### Step 3: Simplify the equation
When we simplify, we get:
\[ 6x - (-10x) - 10y - 6y + 9 - 25 = 0 \]
This simplifies to:
\[ 6x + 10x - 10y - 6y + 9 - 25 = 0 \]
Combining like terms:
\[ 16x - 16y - 16 = 0 \]

### Step 4: Factor out the common factor
We can factor out 16 from the equation:
\[ 16(x - y - 1) = 0 \]

### Step 5: Write the equation of the common chord
Thus, we have:
\[ x - y - 1 = 0 \]
This can be rearranged to give the equation of the common chord:
\[ x - y = 1 \]

### Final Answer
The equation of the common chord of the circles is:
\[ x - y = 1 \]

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