Lines and Tangents Problem

This afternoon we were working on a more complex geometry problem that is a particularly good example of finding the length of a line when the question doesn’t give us straightforward information with which to solve the problem. Here we have indirect information from a tangent to a circle. We are to find the length of |AB|.

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Our fist issue here is that we don’t know the y value of A or the x value of B. As A is on the y axis we know that x=0, and as B is on the x axis y=0. P is on the |AB|, but we know neither its x or y value. Once we have discovered these we have established a reference on |AB| that will lead us to a solution for both A and B.

P is on the line of the circle and is also the point of the tangent. Thus the first thing we must do is find the gradient or ‘slope’ m1 of the |OP|. To do this we must find the x and y of P. We can do this using some simple trigonometry.

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Knowing that |OP| is also the r of the circle we know that the hypotenuse of the right triangle we have constructed is 5 in length. As the measure of the relevant angle is given as 62° we know also that θ=62. The length of the adjacent side will give us the x value of P and the length of the opposite will give us the y value of P.

Cosineθ = adj=xhyp
Cosine62 = adj=x5
5Cosine62 = x
x = 2.347

And y:

Sineθ = opp=yhyp
Sine62 = opp=y5
5Sine62 = y
y = 4.415

We now know P is the point (2.347, 4.415). What we have now is a reference to the heretofore unknown |AB|, and we can now hit it with the equation of the line y=mx+c. Well, not quite. Before we can do this we will have to find the m2 – that is the slope of the second line, |AB|.

As |AB| is the tangent of |OP|, m2 is the negative reciprocal of m1. This means that we must fist find m1 and then work our n-1 magic on it to get m2.

m1 = Δy/Δx
m1 = (y2-y1)/(x2-x1)
m1 = (4.415-0)/(2.347-0)
m1 = 4.415/2.347
m1 = (44152347)

Therefore:

m2 = -1/m1
m2 = -1/(44152347)
m2 = (-23474415)

Now that we have discovered a reference to a point on |AB|; P which is point (2.347, 4.415), and that we know the slope or m2 of that line, we can use the equation y=mx+c to identify first point A and then point B. We start with A because A is the point where |AB| intersects the y axis, and we know this to be c in the equation of the line. Thus:

y = mx+c
4.415 = (-23474415)(2.347) + c
4.415 – 2.347(-23474415) = c
c = 5.663
∴ A = point (0, 5.663)

Lastly – yes we’re nearly done, to find B we have to play around with the y=mx+c equation because in it the mx functions as a single number – that is the ∑ of m·x. All we want is the value of x. So to isolate x we can do this:

y=mx+c
y-c = mx
∴ (y-c)/m = x

At point B y=0, as we are told in the question. This means that only x is still unknown. This means that all we have to do now is plug the known values into out manipulated equation:

(y-c)/m = x
(0-5.663)/(-23474415) = x
-5.663/(-23474415) = x
x = 10.653
∴ B = point (10.653, 0)

We are now ready to calculate the distance of the |AB|, knowing that A is point (0, 5.663) and B is point (10.653, 0).

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What we need here is our distance formula δ=√[(x2-x1)2+(y2-y1)2], where δ is |AB|. So let’s get that done now and wrap this baby up for the day:

δ=√[(x2-x1)2+(y2-y1)2]
|AB| = √[(10.653-0)2+(0-5.663)2]

|AB| = 12.065

So there we have it peeps, the length of the |AB| is 12.065 (rounded to 3 decimal places). Nothing in this was particularly difficult, but it is ‘complex’ in that we needed to know a few different types of maths and work out the solution over a few steps. I really hope this was useful to you, and if you have any problems or comments please just write them in the Thoughts and Questions section below.

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