The abscisae of A and B are the roots of the equation `x ^(2) + 2ax -b ^(2) =0` and their ordinates are the roots of the equation `y ^(2) + 2 py -q ^(2) =0.` The equation of the circle with AB as diameter is

Answer

To find the equation of the circle with AB as the diameter, we will follow these steps:

### Step 1: Identify the roots of the equations
The abscissae (x-coordinates) of points A and B are the roots of the equation:
\[ x^2 + 2ax - b^2 = 0 \]

Let the roots be \( x_1 \) and \( x_2 \). By Vieta's formulas:
- The sum of the roots \( x_1 + x_2 = -\frac{b}{a} = -2a \)
- The product of the roots \( x_1 \cdot x_2 = \frac{c}{a} = -b^2 \)

The ordinates (y-coordinates) of points A and B are the roots of the equation:
\[ y^2 + 2py - q^2 = 0 \]

Let the roots be \( y_1 \) and \( y_2 \). Again, by Vieta's formulas:
- The sum of the roots \( y_1 + y_2 = -\frac{b}{a} = -2p \)
- The product of the roots \( y_1 \cdot y_2 = \frac{c}{a} = -q^2 \)

### Step 2: Write the general equation of the circle
The general form of the equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]

### Step 3: Expand the equation
Expanding the equation:
\[ (x^2 - (x_1 + x_2)x + x_1x_2) + (y^2 - (y_1 + y_2)y + y_1y_2) = 0 \]

### Step 4: Substitute the values from Vieta's formulas
Substituting the values we found:
- \( x_1 + x_2 = -2a \)
- \( x_1 x_2 = -b^2 \)
- \( y_1 + y_2 = -2p \)
- \( y_1 y_2 = -q^2 \)

The equation becomes:
\[ x^2 - (-2a)x - b^2 + y^2 - (-2p)y - q^2 = 0 \]

### Step 5: Simplify the equation
This simplifies to:
\[ x^2 + 2ax - b^2 + y^2 + 2py - q^2 = 0 \]

### Step 6: Rearrange the equation
Rearranging gives us:
\[ x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0 \]

### Final Equation
Thus, the equation of the circle with AB as the diameter is:
\[ x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0 \]

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