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|.


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.


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)


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-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).


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:

|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.


Thoughts and Questions

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s