1.  x^2+y^2=4  is the circle of the radius 2 with the center at the origin (0,0).

2.  (x-4)^2 + (y-3)^2=k^2  is the circle of the radius "k" with the center at the point (4,3).

    Notice that this center lies at the distance 5 =  from the origin.

3.  The condition requires that the circle #1 touches the circle #2.

    It is possible in two cases:

    k = 5 - 2 = 3  (exterior touching),  and'

    k = 5 + 2 = 7  (interior touching).

Since the circles INTERSECT at just 1 point, it obviously means that they only TOUCH each other at a point on the OUTSIDE of their circumferences

The 1st circle: [] is centered at the origin (0, 0), and has a radius of r = 2 ()

The 2nd circle: [] is centered at (4, 3) and has a radius of r = k

Since the circles touch each other at 1 point, and the 2nd circle’s center is (4, 3), it follows that they MUST TOUCH each other in the 1st quadrant 

When joined, the centers of the 2 circles form the 2 radii of the circles, and these radii, combined, happen to form the hypotenuse of a 3-4-5 right triangle. This hypotenuse = 5



Since the circle centered at (0, 0) has a radius of 2, it follows that the circle that’s centered at (4, 3) has a radius of 5 – 2, or 3, so 



The 2nd circle: [], centered at (4, 3), can also be much larger than the circle centered at (0, 0), and so, can have a  (3 + 2 + 2), or 3, 

plus the diameter of the circle centered at (0, 0). This means that the smaller circle [centered at (0, 0)] will be inscribed in the larger circle.



If it's confusing, you'd see a much clearer picture if you draw a diagram, as I did.

It's as easy as that...nothing COMPLEX! 

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