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Let $C$ be a circle of radius $r$ in the plane. Let $p$ be a point in the plane that lies outside of $C$. Show that there are exactly two lines through $p$ that are tangent to $C$.


It is one of those questions that seem very intuitive but very hard to prove for me. How do I show that there are "exactly" two tangent lines? Try to construct a third one but reach a contradiction? I'd appreciate any help. Thanks.

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  • $\begingroup$ Do you know calculus? $\endgroup$ – Kaj Hansen Apr 21 '14 at 2:58
  • $\begingroup$ @Kaj_H yes, I do. It never occurred to me it has anything to do with that though. $\endgroup$ – codeedoc Apr 21 '14 at 3:00
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Method Using Calculus:

Say we are given any circle and any point outside of that circle. WLOG, we can translate the circle to be centered at the origin, and rotate our system so that the point is situated along the $y$-axis.

From here, we claim that we can hit that point with exactly two tangent lines to the circle. Imagine the circle is described by the equation $x^2 + y^2 = r^2$, and consider some point outside of the circle on the $y$-axis, say $(0, b)$.

Differentiating implicitly, we find that $\frac{dy}{dx} = -x/y$. Therefore, the tangent line through a given point on the circle, say $(m, n)$, will be described by the equation $y = (-m/n)x + b$, where $b$ is the $y$-intercept at the desired point. Plugging in to solve for $b$:

$$n = -m^2/n + b$$ $$b = n+ m^2/n$$

We want to find points along our circle that both satisfy the above, and also $x^2 + y^2 = r^2$. Rearranging the above, we get:

$$m^2 + n^2 = nb$$

In order to satisfy our circle's equation, we must take $n$ such that $nb = r^2 \Longrightarrow n = r^2/b$. From here, we have exactly two choices for $m$ (positive or negative), and hence, exactly $2$ possible tangent lines to the circle through the desired point.

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A proof

Let be $s$ a tangent line through $P$. Call $T$ the point in which the line touches $C$. The radius at $T$ is perpendicular to $s$.

So you have to find the points $T$ such that the angle $\angle PTO$ is right (I have called $O$ the center of $C$).

But these points have to lay on the circle $C'$ whose diameter is $OP$. Since $O$ is inside $C$ and $P$ is outside, there are exactly two points of intersection between $C$ and $C'$, which give two tangent lines.

Another proof

Note, as in previous proof, that the tangent line and the corresponding radius are perpendicular. So $POT$ is a right triangle. This means that $$OP^2=r^2+PT^2$$

Sinnce $r$ and $OP$ are fixed, so is $PT$. That is, the points of tangency are at a fixed distance $r'$ from $P$. By triangle inequality, $|r-r'|<OP<r+r'$, from which we can deduce that the circle $C''$ with center at $P$ and radius $r'$ intersects two points on $C$. These are the poinrs of tangency.

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A proof using analytic geometry --

We can place the circle of radius $ \ r \ $ with its center at the origin, so its equation is $ \ x^2 \ + \ y^2 \ = \ r^2 \ $ . We can also pick a point $ \ (C, 0 ) \ $ on the positive $ \ x-$ axis. (This is equivalent to just saying we have some circle and some external point, so we'll define the center of the circle as the "origin" and the line through the center and the external point as the $ \ x-$ axis. So far, this looks like Kaj_H's set-up.

A tangent line to the circle makes contact with it at a point $ \ (X,Y) \ $ , with the coordinates satisfying $ \ X^2 \ + \ Y^2 \ = \ r^2 \ $ . The radius from the origin to that point has slope $ \ m \ = \ \frac{Y}{X} \ $ . The tangent line is perpendicular to that radius, so its slope is $ \ m' \ = \ \frac{1}{m} \ = \ -\frac{X}{Y} \ $ . The equation of the tangent line is then

$$ y \ - \ 0 \ = \ \left( -\frac{X}{Y} \ \right) \ \cdot \ ( \ x - C \ ) \ \ \Rightarrow \ \ y \ = \ \frac{XC}{Y} \ - \ \frac{X}{Y} x \ \ . $$

Now, a tangent point on that line is given by

$$ Y \ = \ \frac{XC}{Y} \ - \ \frac{X}{Y} \cdot X \ \ \Rightarrow \ \ Y^2 \ = \ XC \ - \ X^2 \ \ \Rightarrow \ \ X^2 \ + \ Y^2 \ = \ XC \ \ . $$

[We know this manipulation is "safe" because a tangent line is not going to contact the circle at $ \ Y = 0 \ $ . ]

This means that $ \ XC \ = \ r^2 \ $ ; since we chose $ \ C \ $ to be positive, $ \ X \ $ must be as well. But we also have that $ \ Y^2 \ = \ r^2 \ - \ X^2 \ $ . A tangent line cannot contact the circle at $ \ X = 0 \ $ , as this would require a tangent line of slope zero (we see this from the line equation above). Consequently, $ 0 \ < \ r^2 \ - \ X^2 \ < \ r^2 \ $ , so $ \ Y \ $ has two permissible values; thus, there are two possible tangent lines.

This argument is (unavoidably) related to those offered by ajotatxe and Kaj_H.

[A heuristic "proof": I am standing outside a circular garden, loop of pavement, etc. I see a left-hand edge and a right-hand edge. Lines-of-sight to the edges follow tangent lines.]

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