For what values of a
will the set of equations:

x² + y² = 4 + 2ax - a²
x² = y²

have 4 distinct solutions?

Since x = plus or minus y will be part of any solution, we are looking for values of a which have 2 distinct x solutions for the following equation in x:
2x^2 - 2ax + (a^2 - 4) = 0

x = [2a +/- sqrt(4a^2 - 8a^2 + 32)] / 4
x = [a +/- sqrt(8 - a^2] / 2

If a^2 = 8, there is only one real x solution, so only 2 real (x,y) solutions.
If a^2 < 8, then there are 2 real x solutions and 4 real (x,y) solutions.
If a^2 > 8, then there are 2 imaginary x solutions and 4 imaginary (x,y) solutions.

If: -2sqrt(2) < a < 2sqrt(2) then there are 4 real solutions
If: a < -2sqrt(2)  OR  a > 2sqrt(2) then there are 4 imaginary solutions
And if a is imaginary, I think there will also be 4 imaginary solutions.

So 4 distinct solutions for all values of a except for +/- 2sqrt(2)

Edited on February 20, 2023, 10:43 am

Posted by Larry on 2023-02-20 09:45:26