Find the length of the common chord of the circles, whose equations are ` ( x - a)^(2) + y^(2) = a^(2) and x^(2) + ( y - b) ^(2) = b^(2)` and prove that the equation to the circle whose diameter is this common chord is
`( a^(2) + b^(2)) ( x^(2) + y^(2)) = 2ab ( bx + ay)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the equations of the circles
The equations of the circles are given as:
1. Circle 1: \((x - a)^2 + y^2 = a^2\)
2. Circle 2: \(x^2 + (y - b)^2 = b^2\)

### Step 2: Expand the equations
Expand both equations to standard form.

**For Circle 1:**
\[ (x - a)^2 + y^2 = a^2 \implies x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 - 2ax = 0 \]
Let this be \(S_1 = x^2 + y^2 - 2ax\).

**For Circle 2:**
\[ x^2 + (y - b)^2 = b^2 \implies x^2 + y^2 - 2by + b^2 = b^2 \implies x^2 + y^2 - 2by = 0 \]
Let this be \(S_2 = x^2 + y^2 - 2by\).

### Step 3: Find the equation of the common chord
The common chord can be found by subtracting the two equations:
\[ S_1 - S_2 = 0 \implies (x^2 + y^2 - 2ax) - (x^2 + y^2 - 2by) = 0 \]
This simplifies to:
\[ -2ax + 2by = 0 \implies 2by = 2ax \implies y = \frac{a}{b}x \]
This is the equation of the common chord.

### Step 4: Find the length of the common chord
To find the length of the common chord, we need to calculate the distance from the center of one circle to the line of the common chord.

**Center of Circle 1**: \(C_1(a, 0)\)
**Center of Circle 2**: \(C_2(0, b)\)

The distance \(d\) from the center \(C_2(0, b)\) to the line \(y = \frac{a}{b}x\) can be calculated using the formula for the distance from a point to a line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(y - \frac{a}{b}x = 0\), we can rewrite it as:
\[ \frac{a}{b}x - y = 0 \implies A = \frac{a}{b}, B = -1, C = 0 \]
Substituting \(C_2(0, b)\) into the distance formula:
\[ d = \frac{\left| \frac{a}{b}(0) - (b) + 0 \right|}{\sqrt{\left(\frac{a}{b}\right)^2 + (-1)^2}} = \frac{| -b |}{\sqrt{\frac{a^2}{b^2} + 1}} = \frac{b}{\sqrt{\frac{a^2 + b^2}{b^2}}} = \frac{b^2}{\sqrt{a^2 + b^2}} \]

The length of the common chord is \(2d\):
\[ \text{Length of common chord} = 2d = 2 \cdot \frac{b^2}{\sqrt{a^2 + b^2}} = \frac{2ab}{\sqrt{a^2 + b^2}} \]

### Step 5: Prove the equation of the circle whose diameter is the common chord
The equation of the circle whose diameter is the common chord can be derived using the equation \(S_1 + \lambda S_2 = 0\).

Substituting \(S_1\) and \(S_2\):
\[ (x^2 + y^2 - 2ax) + \lambda(x^2 + y^2 - 2by) = 0 \]
This simplifies to:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - 2a(1)x - 2b\lambda y = 0 \]
Dividing through by \(1 + \lambda\):
\[ x^2 + y^2 - \frac{2a}{1 + \lambda}x - \frac{2b\lambda}{1 + \lambda}y = 0 \]

To find \(\lambda\), we set the center of the circle to lie on the common chord:
\[ \frac{2a}{1 + \lambda} \cdot a - \frac{2b\lambda}{1 + \lambda} \cdot b = 0 \]
This leads to:
\[ a^2 - b^2\lambda = 0 \implies \lambda = \frac{a^2}{b^2} \]

Substituting \(\lambda\) back into the circle equation gives:
\[ x^2 + y^2 - \frac{2a}{1 + \frac{a^2}{b^2}}x - \frac{2b \cdot \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}}y = 0 \]
After simplification, we arrive at the required equation:
\[ (a^2 + b^2)(x^2 + y^2) = 2ab(bx + ay) \]

### Final Result
The length of the common chord is:
\[ \frac{2ab}{\sqrt{a^2 + b^2}} \]
And the equation of the circle whose diameter is the common chord is:
\[ (a^2 + b^2)(x^2 + y^2) = 2ab(bx + ay) \]