The Domain

The the given fraction cannot have the denominator zero.

Nor can it have the expression inside the square root to be negative.

Thus, we conclude that,

x^2+y^2-4>0

x^2+y^4>4

Which is a circle of radius 2, but all the points are strictly OUTSIDE the circle.

The Range

This is tricker.

Let use write,

z=2/sqrt(x^2+y^2-4) -1

We need to find all possible z, such that the equation above has at least one solution (x,y) in the domain.

Add 1 to both sides,

z+1=2/sqrt(x^2+y^2-4).

If z=-1 then we have zero on left and and expression which cannot be zero on right. Thus, z not = -1.

Since z not = -1 that means z+1 is non-zero. And we can take reciprocals of both sides.

Thus,

1/(z+1)=sqrt(x^2+y^2-4)/2

Multiply both sides by 2,

2/(z+1)=sqrt(x^2+y^2-4)

To have a solution we need that the expression on the left to be positive (because square roots are always positive).

Thus we require that,

2/(z+1)>0 solve inequality z>-1.

Now square both sides,

4/(z+1)=x^2+y^2-4

Add 4 to both sides,

4/(z+1)+4 = x^2+y^2

Now here is the important part.

So far we have shown that z>-1 is the necessary condition for solutions to exists. But is that enought.

Meaning does the equation above solution in x,y?

Well, you might say, just make x=0 and solve for y.

But it is not so simple, because the domain is x^2+y^2>4.

Thus, we are asking can we solve for x,y such that x^2+y^2>4.

Yes, we can!

Because the left hand term is,

4/(z+1)+4>4 because 4/(z+1)>0.

Thus, z>-1 is the range.