How to find $k$ such that the line $y=x-2-k$ is tangent to the circle given by $x^2+(y+2)^2=4$? I have the circle  $x^2+(y+2)^2=4$ and the line $y=x-2-k$. How would you find a $k$ value that would allow the second equation to sit tangent to the circle? There should, in theory, be only two solutions. 
I can't make sense of this problem as you can't equate these problems to find a point. Thinking of this problem, I would think of using the discriminate, but I cannot find the proper equation format by manipulating the equation in order to make it fit in the form $ax^2+bx+c$.
Would anyone have any clue on how you may find this value? I've noticed by typing it into a CAS, it partially solves it by stating the domain of k can only be within $2\sqrt2$ or $-2\sqrt2$ which are the solutions to this problem but I would like to understand how it may have equated this domain. 
Thank you for your help.
 A: The line is tangent to the circle iff they intersect at exactly one point. The intersection is given by a quadratic equation, the number of solutions is given by the sign of the discriminant (you want the discriminant to be $0$).
The circle and the line intersect at $x$ given by $x^2+(x-k)^2=4$, i.e. $2x^2-2kx+k^2-4=0$.
There is only one possible $x$ if $\Delta=32-4k^2=0$, that is when $k=\pm2\sqrt{2}$.
Then the coordinates of the tangent points are given by $x=\frac k2, y=-\frac k2-2$.
A: There's another approach with calculus.
The slope of equation $y = x-2-k$ is 1.
Differentiating the equation of circle we get $$\frac{dx^2}{dx} + \frac{d(y+2)^2}{dx} = 0 \\
2x + 2(y+2)\frac{d(y+2)}{dx} = 0\\
x + (y+2)\frac{dy}{dx} = 0\\
\frac {dy}{dx} = \frac{-x}{y+2}$$
This gives the slope of tangent at any point on the circle.
Equating this to 1 we get
$-x=y+2$ substituting it in equation of circle, we get $x=\pm \sqrt 2 \implies y=\mp \sqrt 2 -2$
It means the tangents at point $(\sqrt 2, -\sqrt 2 -2)$ and $(-\sqrt 2, \sqrt 2 -2)$ to the circle have slope 1 which is exactly the slope of the given line.
Substituting these values in the equation $y=x-2-k$, we get 
$$k=\pm 2\sqrt 2$$
