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1.
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Learn the song:
Common tangent pro-ce-dure, EIEIO.
Add your radii at a slant, EIEIO.
With a right angle here! Right angle there!
Here a right angle, there a right angle, everywhere a . . . (5 in all),
Call Mr. Pythagoras on the phone, EIEIO.
[Second verse]
There is only one exception, EIEIO,
But we can use our common sense there. [End of song.]
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2.
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Sample
external tangent problem: Find
the length of belt needed to connect two pulleys whose diameters are 10 and 2
inches if their centers are 8 inches apart.
Solution: First, make a sketch. If the diameters are 10 and 2, the
radii are 5 and 1 as shown. By symmetry, trapezoid ACQP is congruent to
trapezoid TUQP, a fact that will be useful later.
The second step is to add rectangle ACQB in such a way that the segment
joining the centers P and Q is kept as a hypotenuse.
Students often make a mistake on this step, attempting (hopelessly) to make
BCQP be a rectangle. (That is impossible, since then  ACB would be a right triangle with 2 right angles!) All you
have to do to avoid this mistake is to keep the hypotenuse on the line that
joins the centers of the circles. Here is the correct way to add rectangle
ACQB:
Now, look at right triangle BQP, which has a leg of 4 and a hypotenuse of 8.
By the Pythagorean Theorem shortcut,  . However, if  has sides of 4,  , and 8, then must be a 30°-60°-90° triangle, since only a 30°-60°-90°
triangle can have sides in the pattern of x,
 , and 2x.
Therefore,  = 60°.
By the congruence of trapezoids ACQP and TUQP,  = 60° also. Since  = 90° + 30° = 120°,
we know  = 120° also.
Therefore, arc CMU is 240°, forcing minor arc CU to be 120°, or 1/3 of a circle.
(This is important, since the belt connecting the pulleys must cover minor
arc CU.)
Notice that in circle P, minor arc ALT is 120°, forcing major arc TZA to be
240°, or 2/3 of a circle.
By a property of rectangles, AC = BQ = , and by the congruence of trapezoids ACQP and TUQP, TU = also. We now have
all the ingredients for the belt that connects the pulleys:
- AC =
- minor arc CU = 1/3 of the circumference of
circle Q =
- UT =
- major arc TZA = 2/3 of the circumference of
circle P =
Add them all up:  36.895 inches.
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4.
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Sample
internal tangent problem: The
centers of two circles with radii 9 and 11 are 32 units apart. Find the
length of a common internal tangent.
Solution: First, make a diagram
showing the common internal tangent.
Second, add the line joining the centers, Q and R. Also add two perpendicular
radii, and label all known lengths.
Finally, extend  to join a line through Q that is parallel to  .
Do you see that this creates rectangle WTVQ?
Note: As we discussed in class, a right triangle has now
been created with its hypotenuse
along the line that joins the centers. This triangle,  ,
overlaps the rectangle. This is different from what we saw in the external
tangent procedure, in which the triangle and rectangle had a side in common
but did not overlap.
The goal of the problem is to find WT, the length of the common internal
tangent.
By properties of rectangles, TV = WQ = 9. Also note that if we could find QV,
we would have the answer to the problem, since WT = QV. Apply the Pythagorean
Theorem shortcut taught in class to get  .
Since WT = QV by properties of rectangles, final answer =  .
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