The equation of the circle on the common chord of the circles `(x-a)^(2)+y^(2)=a^(2)` and `x^(2)+(y+b)^(2)=b^(2)` as diameter, is

Options

A. `x^(2)+y^(2)=2ab(bx+ay)`
B. `x^(2)+y^(2)=bx+ay)`
C. `(a^(2)+b^(2))(x^(2)+y^(2))=2ab(bx-ay)`
D. `(a^(2)+b^(2))(x^(2)+y^(2))=2(bx+ay)`

Explanation

To find the equation of the circle on the common chord of the circles \((x-a)^2 + y^2 = a^2\) and \(x^2 + (y+b)^2 = b^2\) as diameter, we can follow these steps:

### Step 1: Identify the centers and radii of the circles

1. **Circle 1**: The equation \((x-a)^2 + y^2 = a^2\) has:
- Center: \((a, 0)\)
- Radius: \(a\)

2. **Circle 2**: The equation \(x^2 + (y+b)^2 = b^2\) has:
- Center: \((0, -b)\)
- Radius: \(b\)