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Find the equations of the lines tangent to the circle $x^2+y^2=r^2$ that pass through the point $(a,0)$.

My book explains that the equation of this line is $y=m(x-a)$ and then we make the substitution:

$$x^2+[m(x-a)]^2=r^2\\(1-m^2)x^2-2am^2x+a^2m^2-r^2=0$$

And then it says for that this equation of second degree have only one root, it's discriminant must be zero. And then, this gives us:

$$ \overbrace{4a^2m^4}^{b^2}-4\overbrace{(1+m^2)}^{a}\overbrace{(a^2m^2-r^2)}^{c}=0$$

I don't understand why finding the tangent lines ammount to computing a discriminant and solving for $m$. I know that making that substitution on $x^2+y^2=r^2$ is something that enables us to find the intersections of those curves, but I'm really lost at knowing what is the role the determinant plays here.

Red Banana
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    When a line intersects a circle, it can do so in two points, one (tangent case), or none (no intersection). Correspondingly when you try solving the relevant quadratic equation, you will get two roots, one or none. Do you see how the discriminant plays a role in this determination? – Macavity Jun 21 '15 at 02:13
  • @Macavity Yes. I understood that from the context of the book, but he's speaking about roots, and then it seems that he computes only the discriminant, instead of computing the roots using the quadratic formula. – Red Banana Jun 21 '15 at 02:16
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    He does not need the roots, only the condition that there should be one root, which is that the discriminant is exactly zero. You would need the root if you were asked to find the point of intersection/tangency as well. – Macavity Jun 21 '15 at 02:17
  • @Macavity Now, why the roots do give the points of intersection of the curves? I'm a little lost at explaining this. – Red Banana Jun 21 '15 at 02:18
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    The roots satisfy both the equation of the line and that of the circle, by construction. Hence they are the points of intersection. – Macavity Jun 21 '15 at 02:18
  • In the second equation you wrote, it should be $4(1+m^2)$ instead of $4(1-m^2)$ – Claude Leibovici Jun 21 '15 at 03:47

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