Math help geometry?....?

Question

A chord of a circle is extended so that the external segment of the secant it forms is 4" longer than the internal segment. The tangent of the circle from the extremity of the secant is 6" long. Find to the nearest hundredth of an inch the length of the chord.

Answer 1

I have been amusing myself during a boring meeting by answering what I thought was this question; when I come to put in the answer I see that I was actually answering a slightly different question.  

The question which I worked on I thought had said that the external segment of the secant was 4" long.  I realise now that this was incorrect.

I shall try to work out the answer to the actual question; but in the mean time, I submit the answer to the slightly different question as it may be of interest.

Let the centre of the circle be O, and its radius R.

The chord is AB, and it is extended outside the circle to point C.

The tangent is CD; we are given that  BC = 4",  DC = 6".

M is the mid-point of chord  AB, angle MOB is θ

In triangle  MOB,  MB  =  R sinθ,  OM  =  R cosθ

In triangle OMC,  MC  =  MB + BC  =  Rsinθ + 4

and  OC²  =  OM² + MC²  =  R²cos²θ + (Rsinθ + 4)²

In triangle ODC,  OC²  =  OD² + DC²  =  R² + 6²

(if making a sketch diagram, note that OM and OD are not the same line)

Equating these two expressions for OC²,

R²cos²θ + (Rsinθ + 4)²  =  R²  +  6²

R²cos²θ + R²sin²θ + 8Rsinθ + 16  =  R² + 6²

R²(sin²θ + cos²θ) + 8Rsinθ + 16  =  R² + 36

So  8Rsinθ +16  =  36  (since sin²θ + cos²θ =1)

8Rsinθ  =  20

sinθ  =  20 / 8R  =  5 / 2R

Length of chord  AB  =  2MB  =  2Rsinθ  =  2R.5 / 2R  =  5

From which we can see that the chord length is not dependent upon the radius of the circle: there is an infinite number of circles which would fulfil the given conditions.

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If we substitute  s for the length of the secant exterior to the circle (instead of  4"), and substitute t for the length of the tangent (instead of 6") and repeat the treatment above, we get

sinθ  =  (t² - s²) / 2Rs    and chord length  =  (t² - s²) / s

And it is easy to see that there is a minimum size of circle, when the chord length = 2R (that is, the  chord is a diameter of the circle).

A quick diagram shows directly that, in this case,  (R + s)²  =  R²  +  t²

Hence  R² + 2Rs + s²  =  R² + t²

2Rs + s²  =  t²,  or  2R  =  (t² - s²) / s

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So, now for the actual question.

Using the same notation, this time we have  BC  =  AB + 4

AB  =  2MB ;  AC  =  2AB + 4  =  4MB + 4

MC  =  AC - MB  =  3MB + 4

As before,  OM  =  Rcosθ,   MB  =  Rsinθ

OC²  =  OD²  +  DC²  =  R²  +  6²

Also,  OC²  =  OM²  +  MC²  =  (Rcosθ)²  +  (3Rsinθ + 4)²

So  R² +  6²  =  (Rcosθ)²  +  (3Rsinθ + 4)²

R²  +  6²  =  (Rcosθ)²  +  (3Rsinθ + 4)²

R²  +  6²  =  R²cos²θ  +  9R²sin²θ  +  24Rsinθ  +  16

R²  +  6²  =  R²(cos²θ   +  sin²θ)  +  8R²sin²θ  +  24Rsinθ  +  16

8R²sin²θ  +  24Rsinθ  -  20  =  0

Let  X = Rsin²θ, then the equation is  8X²  +  24X  -  20  =  0

or,  2X²  +  6X  -  5  =  0

Therefore  X  =  (- 6 ±√76) / 4

X  =  - 3.67945  or  0.67945

(Drop the negative value), so  Rsinθ = 0.67945

And the chord length  =  2Rsinθ  =  1.3589

The external segment of the secant  =  2Rsinθ + 4  =  5.3589

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