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

Answer

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

### Step 1: Identify the centers and radii of the circles
- The first circle \((x-a)^2 + y^2 = a^2\) has:
- Center \(C_1 = (a, 0)\)
- Radius \(R_1 = a\)

- The second circle \(x^2 + (y-b)^2 = b^2\) has:
- Center \(C_2 = (0, b)\)
- Radius \(R_2 = b\)

### Step 2: Find the equation of the common chord
The common chord can be found using the equation \(S_1 - S_2 = 0\), where \(S_1\) and \(S_2\) are the equations of the circles.

1. The equation of the first circle:
\[ S_1 = (x-a)^2 + y^2 - a^2 \]
Expanding this gives:
\[ S_1 = x^2 - 2ax + a^2 + y^2 - a^2 = x^2 - 2ax + y^2 \]

2. The equation of the second circle:
\[ S_2 = x^2 + (y-b)^2 - b^2 \]
Expanding this gives:
\[ S_2 = x^2 + y^2 - 2by + b^2 - b^2 = x^2 + y^2 - 2by \]

3. Setting \(S_1 - S_2 = 0\):
\[ (x^2 - 2ax + y^2) - (x^2 + y^2 - 2by) = 0 \]
Simplifying this gives:
\[ -2ax + 2by = 0 \quad \Rightarrow \quad 2by = 2ax \quad \Rightarrow \quad y = \frac{a}{b}x \]

### Step 3: Find the coordinates of the intersection points
To find the intersection points of the two circles, we can substitute \(y = \frac{a}{b}x\) into one of the circle equations. Let's use the first circle's equation:

1. Substitute into \((x-a)^2 + y^2 = a^2\):
\[ (x-a)^2 + \left(\frac{a}{b}x\right)^2 = a^2 \]
Expanding this gives:
\[ (x-a)^2 + \frac{a^2}{b^2}x^2 = a^2 \]
\[ x^2 - 2ax + a^2 + \frac{a^2}{b^2}x^2 = a^2 \]
\[ (1 + \frac{a^2}{b^2})x^2 - 2ax = 0 \]
Factoring out \(x\):
\[ x\left((1 + \frac{a^2}{b^2})x - 2a\right) = 0 \]
Thus, \(x = 0\) or \((1 + \frac{a^2}{b^2})x = 2a\), giving:
\[ x = \frac{2a}{1 + \frac{a^2}{b^2}} = \frac{2ab^2}{a^2 + b^2} \]

2. Now find \(y\) using \(y = \frac{a}{b}x\):
\[ y = \frac{a}{b} \cdot \frac{2ab^2}{a^2 + b^2} = \frac{2a^2b}{a^2 + b^2} \]

### Step 4: Find the equation of the circle with diameter OA
The points \(O(0, 0)\) and \(A\left(\frac{2ab^2}{a^2 + b^2}, \frac{2a^2b}{a^2 + b^2}\right)\) are the endpoints of the diameter. The equation of the circle can be given by:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]
Where \(x_1 = 0\), \(y_1 = 0\), \(x_2 = \frac{2ab^2}{a^2 + b^2}\), and \(y_2 = \frac{2a^2b}{a^2 + b^2}\).

Thus, the equation becomes:
\[ x\left(x - \frac{2ab^2}{a^2 + b^2}\right) + y\left(y - \frac{2a^2b}{a^2 + b^2}\right) = 0 \]

### Final Equation
The final equation of the circle is:
\[ x^2 + y^2 - \frac{2ab^2}{a^2 + b^2}x - \frac{2a^2b}{a^2 + b^2}y = 0 \]