Two circles of equal radius of 5 units have their centres at the origin and the point `(2, -3).` Equation of the common chord of these circles is

Options

A. `4x- 6y -13 =0`
B. `2x - 3y + 13=0`
C. `4x-6y + 12=0`
D. `2x -3y-12=0`

Answer

To find the equation of the common chord of the two circles with centers at the origin (0, 0) and (2, -3), both having a radius of 5 units, we will follow these steps:

### Step 1: Write the equations of both circles.

1. **Circle 1 (C1)**:
- Center: (0, 0)
- Radius: 5
- Equation:
\[ (x - 0)^2 + (y - 0)^2 = 5^2 \]
Simplifying this gives:
\[ x^2 + y^2 = 25 \quad \text{(Equation 1)} \]

2. **Circle 2 (C2)**:
- Center: (2, -3)
- Radius: 5
- Equation:
\[ (x - 2)^2 + (y + 3)^2 = 5^2 \]
Expanding this gives:
\[ (x^2 - 4x + 4) + (y^2 + 6y + 9) = 25 \]
Combining terms leads to:
\[ x^2 + y^2 - 4x + 6y + 13 = 25 \]
Rearranging gives:
\[ x^2 + y^2 - 4x + 6y - 12 = 0 \quad \text{(Equation 2)} \]

### Step 2: Find the equation of the common chord.

To find the equation of the common chord, we can use the formula:
\[ S_1 - S_2 = 0 \]
where \(S_1\) is the equation of Circle 1 and \(S_2\) is the equation of Circle 2.

1. **Substituting the equations**:
- From Equation 1: \(S_1 = x^2 + y^2 - 25\)
- From Equation 2: \(S_2 = x^2 + y^2 - 4x + 6y - 12\)

2. **Subtracting the two equations**:
\[ S_1 - S_2 = (x^2 + y^2 - 25) - (x^2 + y^2 - 4x + 6y - 12) = 0 \]
Simplifying this gives:
\[ -25 + 4x - 6y + 12 = 0 \]
Rearranging leads to:
\[ 4x - 6y - 13 = 0 \]

### Final Equation of the Common Chord:
The equation of the common chord is:
\[ 4x - 6y - 13 = 0 \]