The abscissa of the two points A and B are the roots of the equation `x^2+2a x-b^2=0` and their ordinates are the roots of the equation `x^2+2p x-q^2=0.` Find the equation of the circle with AB as diameter. Also, find its radius.

Answer

To solve the problem, we need to find the equation of the circle with diameter AB, where A and B are points defined by the roots of two quadratic equations.

### Step-by-step Solution:

1. **Identify the coordinates of points A and B:**
- The abscissas (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, we have:
\[ x_1 + x_2 = -2a \quad \text{(sum of roots)} \]
\[ x_1 x_2 = -b^2 \quad \text{(product of roots)} \]

- 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:
\[ y_1 + y_2 = -2p \quad \text{(sum of roots)} \]
\[ y_1 y_2 = -q^2 \quad \text{(product of roots)} \]

2. **Find the equation of the circle with AB as diameter:**
- The general equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]
- Expanding this, we get:
\[ x^2 - (x_1 + x_2)x + x_1 x_2 + y^2 - (y_1 + y_2)y + y_1 y_2 = 0 \]
- Substituting the values from Vieta's formulas:
\[ x^2 + 2ax + (-b^2) + y^2 + 2py + (-q^2) = 0 \]
- This simplifies to:
\[ x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0 \]

3. **Identify the center and radius of the circle:**
- The center \( (h, k) \) of the circle can be found from the equation:
\[ x^2 + y^2 + 2ax + 2py - (b^2 + q^2) = 0 \]
The center is given by:
\[ h = -a, \quad k = -p \]
- To find the radius \( r \), we use the formula:
\[ r = \sqrt{h^2 + k^2 - c} \]
where \( c = -(b^2 + q^2) \). Thus:
\[ r = \sqrt{(-a)^2 + (-p)^2 + (b^2 + q^2)} = \sqrt{a^2 + p^2 + b^2 + q^2} \]

### Final Results:
- The equation of the circle is:
\[ x^2 + y^2 + 2ax + 2py - b^2 - q^2 = 0 \]
- The radius of the circle is:
\[ r = \sqrt{a^2 + p^2 + b^2 + q^2} \]