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)`

Answer

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: Write the equations of the circles in standard form
The first circle is given by:
\[ (x-a)^2 + y^2 = a^2 \]
Expanding this, we get:
\[ x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 - 2ax = 0 \]

The second circle is given by:
\[ x^2 + (y-b)^2 = b^2 \]
Expanding this, we get:
\[ x^2 + y^2 - 2by + b^2 = b^2 \implies x^2 + y^2 - 2by = 0 \]

### Step 2: Find the equation of the common chord
The equation of the common chord can be found by subtracting the two equations:
\[ (x^2 + y^2 - 2ax) - (x^2 + y^2 - 2by) = 0 \]
This simplifies to:
\[ -2ax + 2by = 0 \implies 2by = 2ax \implies Ax = By \]
Where \(A = a\) and \(B = b\).

### Step 3: Express \(x\) in terms of \(y\)
From the equation \(Ax = By\), we can express \(x\) as:
\[ x = \frac{b}{a}y \]

### Step 4: Substitute \(x\) in one of the circle equations
We can substitute \(x\) into the second circle's equation:
\[ x^2 + (y-b)^2 = b^2 \]
Substituting \(x = \frac{b}{a}y\):
\[ \left(\frac{b}{a}y\right)^2 + (y-b)^2 = b^2 \]
This simplifies to:
\[ \frac{b^2}{a^2}y^2 + (y^2 - 2by + b^2) = b^2 \]
Combining terms gives:
\[ \left(\frac{b^2}{a^2} + 1\right)y^2 - 2by = 0 \]

### Step 5: Factor the equation
Factoring out \(y\):
\[ y\left(\left(\frac{b^2}{a^2} + 1\right)y - 2b\right) = 0 \]
This gives us two solutions: \(y = 0\) or \(\left(\frac{b^2}{a^2} + 1\right)y - 2b = 0\).

### Step 6: Solve for \(y\)
From \(\left(\frac{b^2}{a^2} + 1\right)y - 2b = 0\):
\[ y = \frac{2b}{\frac{b^2}{a^2} + 1} = \frac{2ab^2}{a^2 + b^2} \]

### Step 7: Substitute \(y\) back to find \(x\)
Substituting \(y\) back into \(x = \frac{b}{a}y\):
\[ x = \frac{b}{a} \cdot \frac{2ab^2}{a^2 + b^2} = \frac{2b^3}{a^2 + b^2} \]

### Step 8: Write the equation of the circle
The diameter endpoints are \((0, 0)\) and \(\left(\frac{2b^3}{a^2 + b^2}, \frac{2ab^2}{a^2 + b^2}\right)\). The equation of the circle with these endpoints is:
\[ (x - 0)\left(x - \frac{2b^3}{a^2 + b^2}\right) + (y - 0)\left(y - \frac{2ab^2}{a^2 + b^2}\right) = 0 \]
This simplifies to:
\[ x^2 + y^2 - \frac{2b^3}{a^2 + b^2}x - \frac{2ab^2}{a^2 + b^2}y = 0 \]

### Final Equation
The final equation of the circle is:
\[ (a^2 + b^2)(x^2 + y^2) = 2ab(bx + ay) \]