Abscissas and ordinates of points A and B are roots of
equation `x^(2) + 2ax - b^(2) = 0 " and " y^(2) + 2py - q^(2) = 0 `
respectively. Equation of circle with AB as a diameter is

Options

A. `x^(2) + y^(2) + 2ax + 2py + b^(2) + q^(2) = 0 `
B. `x^(2) + y^(2) - 2ax - 2py - b^(2) - q^(2) = 0 `
C. `x^(2) + y^(2) - 2ax + 2py - b^(2) -q^(2) = 0`
D. `x^(2) + y^(2) - 2ax - 2py + b^(2) + q^(2) = 0 `

Answer

To find the equation of the circle with points A and B as the diameter, we will follow these steps:

### Step 1: Identify the roots of the equations
The abscissas (x-coordinates) of point A and point B are the roots of the equation:
\[ x^2 + 2ax - b^2 = 0 \]
Using the quadratic formula, the roots \( x_1 \) and \( x_2 \) can be found as follows:
\[ x_1 + x_2 = -\frac{b}{a} = -2a \]
\[ x_1 x_2 = \frac{c}{a} = -b^2 \]

Similarly, the ordinates (y-coordinates) of point A and point B are the roots of the equation:
\[ y^2 + 2py - q^2 = 0 \]
Using the quadratic formula, the roots \( y_1 \) and \( y_2 \) can be found as follows:
\[ y_1 + y_2 = -\frac{b}{a} = -2p \]
\[ y_1 y_2 = \frac{c}{a} = -q^2 \]

### Step 2: Write the equation of the circle
The general equation of a circle with diameter endpoints (x1, y1) and (x2, y2) is given by:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]

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

Substituting the sums and products of the roots:
\[ x^2 - (-2a)x - b^2 + y^2 - (-2p)y - q^2 = 0 \]

### Step 4: Simplify the equation
This simplifies to:
\[ x^2 + 2ax + y^2 + 2py - (b^2 + q^2) = 0 \]

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

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