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Two Problems on Tangents

```
Date: 07/09/98 at 21:22:00
From: Kimberly Wah
Subject: 2 tangent segments from a point to a circle

There are two tangent segments from a point to a circle. Say that the
angle that the two tangents form is 80 degrees. How would a person go
about proving that the arc closest to the angle formed by the two
tangents and that angle are supplementary? So therefore the arc is
100. But how do we prove that?
```

```
Date: 07/09/98 at 22:25:44
From: Doctor Pete
Subject: Re: 2 tangent segments from a point to a circle

Hi,

To see why this is true, draw three lines.

First draw the line that goes through the center of the circle and the
given point. This line bisects the angle formed by the two tangents.
(Why?)

Now, draw two radii, from the center of the circle to each of the two
points of tangency on the circle. Each radius is perpendicular to the
tangent line it touches. (Again, why?)

So we have two congruent right triangles. Since the sum of the angles
of each triangle is 180, and the triangle is right, it follows that
half the angle of the intercepted arc is complementary to half the
angle formed by the tangent lines. Multiplying by 2 gives the desired
relationship.

Here is a picture for your reference:

- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 07/10/98 at 11:37:16
From: Kimberly Wah
Subject: Re: 2 tangent segments from a point to a circle

Thank you very much, Dr. Math. Actually, I was playing around with the
lines and things last night and I thought I got the answer but I
wasn't sure until this morning when I got your email. I drew the 2
radii to the points where the tangents met the circle, and I knew they
were perpendicular. Since they were perpendicular, and I had put the
line down the center of the circle to the point, I noticed that they
formed 2 right triangles. Since they were right triangles, the other 2
sides had to equal 90 degrees and since there were 2 triangles, they
must equal 180 degrees. Thanks!
```

```
Date: 07/11/98 at 13:19:00
From: Kimberly Wah
Subject: Re: 2 tangent segments from a point to a circle

Dr. Pete,

I have another question for you. There is one circle with O at the
center. PT is a tangent to circle O. PN intersects circle O at J
coming to a point N in the circle. And then ON is drawn, but it isn't
a radius. TP = 12, PJ = 8, JN = 4, and ON = 4. Find the radius of
circle O.

It looks to me as if I could put an auxiliary line at TN and make an
isosceles triangle, but it doesn't really help me much.

Thanks,
Kimberly Wah
```

```
Date: 07/11/98 at 20:43:25
From: Doctor Pete
Subject: Re: 2 tangent segments from a point to a circle

Hi,

This problem that you gave me was quite a bit more difficult than the
last one! I did find a solution; however, I had to use some
trigonometry, in particular the Law of Cosines. Since you're in
geometry I don't know if you're familiar with this theorem, so I'll
describe it first.

Suppose you have a triangle ABC, with sides of length a, b, c (so the
angle at A subtends the side of length a, etc.) Then:

c^2 = a^2 + b^2 - 2ab Cos(C)

that is, the square of c is equal to the square of a plus the square
of b minus twice ab times the cosine of the angle at C. From this you
can see that if ABC is a right triangle, with C = 90 degrees, then
Cos(C) = 0, and we obtain:

c^2 = a^2 + b^2

which is the Pythagorean Theorem. So the Law of Cosines is really a
generalization of the Pythagorean Theorem for non-right-angled
triangles. If you're not familiar with function "cosine," don't worry
about it. Just remember it's a quantity that depends on angle.

Now, the question is, how do we use the Law of Cosines? To help, I've
drawn the diagram you described above:

Now, consider triangle OTP. Angle OTP is right, and OT = r, TP = 12,
so by the Pythagorean Theorem, OP = sqrt(r^2 + 12^2), the square root
of r squared plus 144. Then consider triangles OJP and ONP, and, in
particular, the angle ONP (which I will abbreviate as simply angle P).
The Law of Cosines applied to triangle OJP gives:

(OJ)^2 = (OP)^2 + (JP)^2 - 2(OP)(JP) Cos(P)

and since OJ = r, JP = 8, and we found OP above, this gives:

r^2 = (r^2 + 12^2) + 8^2 - 16(sqrt(r^2 + 12^2)) Cos(P)

and some cancelling and rearranging gives:

Cos(P) = 13/sqrt(r^2 + 12^2)

But now consider triangle ONP, which also shares angle P with OJP. So
the Law of Cosines gives in this case:

(ON)^2 = (OP)^2 + (NP)^2 - 2(OP)(NP) Cos(P)

and with ON = 4, NP = 12, we obtain (with some algebra):

Cos(P) = (r^2 + 272)/(24(sqrt(r^2+12^2)))

Combining these two results, and cancelling the common denominator of
sqrt(r^2 + 12^2), we finally obtain:

13 = (r^2 + 272)/24
r^2 = 40.

Since r is positive, r = 2 sqrt(10), which is approximately 6.3246.

I could do geometrically, such as using isosceles triangles or
similarities,  but I soon realized that I couldn't figure out how to
relate the lengths to the angles. It became quite obvious that the
lengths were the key, not so much the angles. But I didn't want to
resort to trigonometry - perhaps there is a purely geometrical
solution to the problem. Often many problems in algebra and
trigonometry have purely geometrical solutions (for example, the
Pythagorean Theorem), but the reasoning is usually more complicated.
Here I used a fairly well-known theorem which proves to be very
powerful in relating the angles of a triangle to its sides.

- Doctor Pete, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 07/11/98 at 22:29:11
From: Kimberly Wah
Subject: Re: 2 tangent segments from a point to a circle

Thanks. I'm going to try to see if I can understand this problem.
Yes, we've learned about trigonometry and the Pythagorean theorem.
Thanks again.
```
Associated Topics:
High School Conic Sections/Circles
High School Euclidean/Plane Geometry
High School Geometry
High School Trigonometry

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