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A circle with a radius of $2$ units has its center at $(0,0)$. A circle with a radius of $7$ units has its center at $(15,0)$. A line tangent to both circles intersects the $x$-axis at $(x,0)$. What is the value of $x$? Express your answer as a common fraction.

My problem with this question is that there are $4$ such tangent lines, so how do I know which one to pick?

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    $\begingroup$ If a single answer is expected, the wording of the question is unfortunate. True, there are only $2$ possible answers, not $4$. But $2\gt 1$. $\endgroup$ – André Nicolas Dec 21 '15 at 17:28
  • $\begingroup$ Are you sure there are four? I haven't drawn a picture, but I'm imagining only two. $\endgroup$ – pjs36 Dec 21 '15 at 17:29
  • $\begingroup$ There are two internally tangent lines and two externally tangent lines, correct? And if this question is wrong, that would be bad since this question was given on a national competition. $\endgroup$ – Puzzled417 Dec 21 '15 at 17:31
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    $\begingroup$ @basket There are four lines (see the left figure in this picture), but since this is all symmetrical about the $x$-axis, the two blue lines will intersect the $x$-axis at the same point, and so will the two red ones. $\endgroup$ – Arthur Dec 21 '15 at 17:35
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    $\begingroup$ @basket They are tangent to both circles. What about them makes them not valid? And I agree with André Nicolas, this is a poorly worded problem. $\endgroup$ – Arthur Dec 21 '15 at 17:43
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Edit: a quick search (using the information in the comments) I get the feeling that this question deals with problem 21 of this 2008 Mathcounts competition. If so, then I must say I don't see what the problem is. The question comes with a diagram. This diagram clearly shows which tangent line we're considering. So then this question is not so poor at all.

Edit 2: If I had been paying attention, I would have noticed that the document in the link, contains the solutions to the problems. Could not find the actual question, so it might still be a quite poorly worded question.


Original answer: The question as mentioned in the OP is poorly worded. There are four lines tangent to both circles. They intersect the $x$-axis in two different points. See this geogebra sketch:

enter image description here

Not that the OP asks, but both points of intesection are fairly easy to find, via similar triangles.

N.B. I feel inclined to add that the intersection with the positive $x$-axis is probably the one the question wants, since it says "Express your answer as a common fraction." This is not someting one would add to the question when the answer is $-6$.

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  • $\begingroup$ The other point is where $x = -6$. Thus we have two conflicting answers. $\endgroup$ – Puzzled417 Dec 21 '15 at 18:57
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    $\begingroup$ And that is why this is a poor question. It has more than one correct answer whilst it implies there is only one. $\endgroup$ – gebruiker Dec 21 '15 at 19:01
  • $\begingroup$ The solution you linked quotes "to the right of the origin". Even without the diagram, this makes the problem unambiguous. $\endgroup$ – Corellian Dec 21 '15 at 19:37
  • $\begingroup$ However, the omission of the above phrase certainly allows two solutions. I'm curious where @Puzzled417 encountered his version of the question. $\endgroup$ – Corellian Dec 21 '15 at 19:38
  • $\begingroup$ @Brody you are right, I had left that out by accident. Now the question is clear, right? $\endgroup$ – Puzzled417 Dec 21 '15 at 20:33
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So there is solution using the $Ax+By+C=0$ notation for lines. Given a circle with center $(x_c,y_c)$ and radius $r$ the all the possible tangents are given by the line:

$$ x (\cos\psi) + y (\sin \psi) -(x_c \cos\psi + y_c \sin \psi + r) = 0 $$

In fact, every scalar multiple of the $(A,B,C)$ coefficients also describe the same line and we are going to use this fact below to match the tangent lines for the two circles.

For the two circles the two general tangent lines are:

$$\begin{align} x (\cos \psi_1) + y ( \sin \psi_1) - 2 & =0 \\ x (\cos \psi_2) + y ( \sin \psi_2) - (7+15 \cos \psi_2) & = 0 \end{align}$$

If we multiply the first equation with $\lambda = \frac{7}{2} + \frac{15}{2} \cos\psi_2$ and then subtract it from the second equation you will get

$$ x \left(\cos\psi_2 - \lambda \cos\psi_1 \right) + y \left( \sin\psi_2 - \lambda \sin\psi_1 \right) = 0 $$

This must be true, regardless which point $(x,y)$ along the tangent line is used. So the above is alike a set of equations for the two angles

$$\begin{align} \cos \psi_2 & = \lambda \cos \psi_1 \\ \sin \psi_2 & = \lambda \sin \psi_1 \end{align}$$

This has two solutions, $\lambda=1$ and $\lambda=-1$. The first one gives the outside tangents, and the second the inside tangents. So the two tangents are found by $$ \cos \psi_2 = -\frac{1}{3} \\ \cos \psi_2 = - \frac{3}{5} $$

Solving the second tangent line equation for $x$ using $y=0$ and the angles above gives us

$$ x= \frac{7}{\cos \psi_2} + 15 = \begin{cases} -6 & \lambda=1 \\ \frac{10}{3} & \lambda=-1 \end{cases} $$

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