(x-a)^2 + y^2 = 4

x^2 - y^2 = 0

Then its easier to see the first equation is a circle of radius 2 with its center of the y-axis and the second equation is a degenerate hyperbola which consists of the two lines y=x and y=-x.

So now to have four distinct solutions then each of the two lines must intersect the circle at two distinct points each.

There are four values of a to make note of: a=2sqrt(2) and a=-2sqrt(2) are the values of a where the lines are tangent to the circle; a=2 and a=-2 are the values of a where the lines' intersection is also on the circle.

If we are to consider complex values of a then a is any value except the four values {+/-2sqrt(2), +/-2} then we will have four distinct solutions.  

If we limit to just real values then we will have four distinct solutions when a is in the interval (-2sqrt(2), 2sqrt(2)) with exception of a=+/-2.