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Say we have two circles $C_1$ and $C_2$ with radii $r_1$ and $r_2$, respectively. Let their centers be $d$ units apart. There are 4 bitangents, two outer and two inner.

Examine the intersection of an inner bitangent with an outer one, near $C_1$. I'm looking for the value of the angle in the quadrant that contains $C_2$ in terms of $r_1$, $r_2$, and $d$.

Anyone have a reference (or a solution) for this? I'm sure it's been looked at before, but I don't know where to look.

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I'm going to assume $r_2>r_1$, as in my diagram. I haven't checked to see whether this is necessary for the rest of my work or not. This might get confusing, so let me start with a rough outline. $PX_2C_2I_2$ is a quadrilateral with right angles at $X_2$ and $I_2$ and the desired angle is the interior angle of this quadrilateral at $P$. I'm going to find the measure of the internal angle at $C_2$ by finding the measures of $\angle I_2C_2C_1$ and $\angle X_2C_2C_1$.

Let's start with $\angle X_2C_2C_1$. Consider $\triangle C_1C_2R_x$. It has a right angle at $R_x$, $C_1C_2=d$, and $C_2R_x=r_2-r_1$, so $$\cos(\angle X_2C_2C_1)=\frac{C_2R_x}{C_1C_2}=\frac{r_2-r_1}{d}$$ and $$\angle X_2C_2C_1=\arccos\left(\frac{r_2-r_1}{d}\right).$$

Now, turn to $\triangle C_1C_2R_i$. It has right angle at $R_i$, $C_1C_2=d$, and $C_2R_i=r_1+r_2$, so $$\cos(\angle I_2C_2C_1)=\frac{C_2R_i}{C_1C_2}=\frac{r_1+r_2}{d}$$ and $$\angle I_2C_2C_1=\arccos\left(\frac{r_1+r_2}{d}\right).$$

Thus, $$\angle X_2C_2I_2=\arccos\left(\frac{r_2-r_1}{d}\right)+\arccos\left(\frac{r_1+r_2}{d}\right)$$ and $$\angle X_2PI_2=180°-\arccos\left(\frac{r_2-r_1}{d}\right)-\arccos\left(\frac{r_1+r_2}{d}\right).$$

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The last formula simplifies a bit if we note that $\pi/2-\arccos\,x=\arcsin\,x$... – J. M. Jul 22 '11 at 15:53
@J. M.: Ahh, yeah. Looking at it now, I can see how I could have gotten straight to the arcsine version of the formula geometrically, but it would've been a bit more obscure to get there purely geometrically than to apply your simplification. (Because of the parallel lines, the target angle is congruent to $\angle R_iC_1R_x$.) – Isaac Jul 22 '11 at 15:56

Its useful to make use of the duality of projective geometry which interchanges points and lines in the
plane. The dual of a smooth curve in projective geometry is the collection of tangents to the curve. The projective dual to a conic section is a conic section. Conic sections have degree 2 so two conic sections intersect in 2*2=4 points. Thus there are always 4 bitangents to a pair of conic sections.

We have to take into account complex intersections, intersections at infinity, and the multiplicity of a intersections to account for all 4 bitangents. For instance the x-axis counts as 2 bitangents to the parabolas y=x^2 and y=2*x^2. The line at infinity counts as the other 2 bitangents.

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This is interesting information, but how does it answer the question at hand? – Rahul Jan 2 '13 at 20:42
How do we know that two circles have just four bitangents? If there were a fifth bitangent then the question at hand may not have a unique answer. Fortunately there are only four bitangents. – Scott Hotton Jan 15 '13 at 4:27

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