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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.

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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 $x=\frac k2, y=-\frac k2-2$.

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    $\begingroup$ @Ryan_DS The coefficients $a,b,c$ in $ax^2+bx+c$ are respectively $2, -2k, k^2-4$, and the discriminant is $b^2-4ac=(-2k)^2-4\times2\times(k^2-4)=4k^2-8k^2+32=32-4k^2$. $\endgroup$ – Jean-Claude Arbaut Apr 17 at 8:54
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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 $$k=\pm 2\sqrt 2$$

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  • $\begingroup$ I like the approach, but isn't the differentiated equation of the circle $(2-x)/y$? $\endgroup$ – Ryan_DS Apr 18 at 2:38
  • $\begingroup$ @Ryan_DS It isn't $(2-x)/y$. I edited the answer to include the complete differentiation process. $\endgroup$ – dssknj Apr 19 at 10:51

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