First I'll approach it graphically to see what
to expect; then I'll do it algebraically

The graph of  is this parabola:



The graph of  is this ellipse:



Put them together on one graph:

The graph of  is this ellipse:



Since these graphs cross in two points, these will be two
real solutions, and since parabolas and circles could cross
as many as 4 times, there will two imaginary solutions as well




The first equation is already solved for , so we substitute
 for  in the second equation:










That does not factor, so we solve that for 
using the quadratic formula:



where , , 

 





So we have two solutions for 

 and 

If we substitute the first value of  in , we get

  

If we substitute the second value of  in , we get

 

The first value of  gives these two solutions for x:

 and 

The second value of gives these two solutions for x:

, ,

but since  is a negative number, it's going to be
imaginary, so we factor  out of it, ,
so we have



and of course 

So we have four solutions for (,)

(, ) = (, ) 
(, ) = (, ) 

These are the points where the curves above cross, and are the two
real solutions.

Here are the two imaginary solutions:

(, ) 
(, )
 
Edwin