Problem Statement:
If the line y = x + 3 intersects the circle with equation x^2 + y^2 = a^2 at points A and B, then what is the equation of the circle with AB as its diameter?

Solution:

Step 1: Find the Intersection Points
We need to find the points of intersection between the line y = x + 3 and the circle x^2 + y^2 = a^2.

Substituting y = x + 3 into the equation of the circle, we get:
x^2 + (x + 3)^2 = a^2
Expanding the equation, we have:
x^2 + x^2 + 6x + 9 = a^2
Combining like terms, we obtain:
2x^2 + 6x + 9 - a^2 = 0

Step 2: Solve the Quadratic Equation
To find the intersection points A and B, we need to solve the quadratic equation 2x^2 + 6x + 9 - a^2 = 0.

Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), where a = 2, b = 6, and c = 9 - a^2, we get:
x = (-6 ± √(6^2 - 4(2)(9 - a^2)))/(2(2))
Simplifying further, we have:
x = (-6 ± √(36 - 4(2)(9 - a^2)))/(4)
x = (-6 ± √(36 - 8(9 - a^2)))/(4)
x = (-6 ± √(36 - 72 + 8a^2))/(4)
x = (-6 ± √(8a^2 - 36))/(4)
x = (-6 ± √4(2a^2 - 9))/(4)
x = (-6 ± 2√(2a^2 - 9))/(4)
x = (-3 ± √(2a^2 - 9))/(2)

Step 3: Find the y-values
To find the y-values of the intersection points A and B, we substitute the x-values obtained in Step 2 into the equation y = x + 3.

For x = (-3 + √(2a^2 - 9))/(2), we have:
y = (-3 + √(2a^2 - 9))/(2) + 3
Simplifying, we get:
y = (-3 + √(2a^2 - 9) + 6)/(2)
y = (3 + √(2a^2 - 9))/(2)

Similarly, for x = (-3 - √(2a^2 - 9))/(2), we have:
y = (3 - √(2a^2 - 9))/(2)

Step 4: Determine the Midpoint
The midpoint of the line segment AB is the center of the circle we want to find. To find the midpoint, we average the x and y coordinates of points A and B.

The x-coordinate of

2