Mrthdrb

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 $2sqrt2$ or $-2sqrt2$ 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.

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 $k=pm2sqrt2$.

Then the coordinates of the tangent points are given by $x=frac k2, y=-frac k2-2$.

There's another approach with calculus.

The slope of equation $y = x-2-k$ is 1.
Differentiating the equation of circle we get $$fracdx^2dx + fracd(y+2)^2dx = 0 \
2x + 2(y+2)fracd(y+2)dx = 0\
x + (y+2)fracdydx = 0\
frac dydx = frac-xy+2$$
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
$$k=pm 2sqrt 2$$