Circle questions -- Secants, chords and tangents

Hello again, just got a few questions pertaining to circles. I'll just post one to start with;

"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).

http://i1137.photobucket.com/albums/...mexam10Q32.png

So solving this.. according to the segments of secants and tangents theorem...

x^{2} = (2r + 353)(353)

x^{2} = (12742 + 353)(353)

x^{2} = 4622535

x = = 2150km (approx.)

Did I do that right?

Re: Circle questions -- Secants, chords and tangents

Quote:

Originally Posted by

**StonerPenguin** Hello again, just got a few questions pertaining to circles. I'll just post one to start with;

"The radius of the earth is approximately 6371 km. If the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x).

http://i1137.photobucket.com/albums/...mexam10Q32.png
So solving this.. according to the segments of secants and tangents theorem...

x

^{2} = (2r + 353)(353)

x

^{2} = (12742 + 353)(353)

x

^{2} = 4622535

x =

= 2150km (approx.)

Did I do that right?

Computation looks right to me. But simply memorizing theorems does not promote understanding.

Let's see how we get that theorem. The radius through a point of tangency and the tangent at the same point are perpendicular.

$Let\ r = length\ of\ radius\ of\ given\ circle,$

$u + r = length\ from\ the\ given\ circle's\ center\ to\ a\ given\ point\ outside\ the\ circle,$

$x = length\ of\ the\ line\ that\ is\ tangent\ to\ the\ given\ circle\ and\ runs\ through\ the\ given\ point.$

By the Pythagorean Theorem, we have:

$(u + r)^2 = x^2 + r^2 \implies u^2 + 2ru + r^2 = x^2 + r^2 \implies x^2 = u^2 + 2ru = u(u + 2r) \implies x = \sqrt{u(u + 2r)}.$

Re: Circle questions -- Secants, chords and tangents

ISS is6724 km above center of earth.At that point the angle between straight down and the visible horizon has a sin of 6371/6724 or 71.35 deg

cos 71.35 =d/6724

d km to horizon =2150 km

Re: Circle questions -- Secants, chords and tangents

Hello, StonerPenguin!

Quote:

The radius of the earth is approximately 6371 km.

If the international space station (ISS) is orbiting 353 km above the earth,

find the distance from the ISS to the horizon (x).

Code:

` o`

|\

| \

352 | \

| \ x

| \

* * * \

* | *\

* | o

* 6371| * *

| *6371

* | * *

* o *

* *

* *

* *

* *

* * *

Note the right triangle.

The equation is: .

Re: Circle questions -- Secants, chords and tangents

Thank you JeffM, bjhopper and Soroban! It's nice to see different perspectives :D and the image drawn in code is really cool.

Here's another question I've had trouble with:

http://i1137.photobucket.com/albums/...mexam10Q15.png

"Explain how you know ≅ given *E* is the center of the circle. (Include theorem numbers.)"

And here's some pertinent theorems;

Quote:

Theorem 10.4

If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.

Quote:

Theorem 10.5

If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.

Quote:

Theorem 10.6

In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.

Obviously from the diagram ≅ by theorem 10.6, but I can't really word this well. Any help? Theorems and proofs are my weakest areas :/

Re: Circle questions -- Secants, chords and tangents

t is a triangle a is an angle

t AED congruent to BEC isosceles t's same legs and equal altitudes

a DEC = 180- 2* 1 / 2 *AED

a AEB 180-2*1/2 * BEC a AED = aBEC

a AEB = a DEC

AB =DC equal arcs = equal chords