Question 1056900
Find the value/s of k for which the circle x^2+y^2=4 and (x-4)^2 + (y-3)^2=k^2 intersect at exactly one point.
<pre>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: [{{{x^2 + y^2 = 4}}}] is centered at the origin (0, 0), and has a radius of r = 2 ({{{sqrt(4)}}})
The 2nd circle: [{{{(x - 4)^2 + (y - 3)^2 = k^2}}}] 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 {{{highlight_green(k = 3)}}}

The 2nd circle: [{{{(x - 4)^2 + (y - 3)^2 = k^2}}}], centered at (4, 3), can also be much larger than the circle centered at (0, 0), and so, can have a {{{highlight_green(matrix(1,3, radius, of, 7))}}} (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!