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’ *m _{1}* 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=x}⁄_{hyp}**

*Cosine62 =*

^{adj=x}⁄5*5Cosine62 = x*

*x = 2.347*

And *y*:

**Sineθ = ^{opp=y}⁄_{hyp}**

*Sine62 =*

^{opp=y}⁄5*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

*m*– that is the slope of the second line, |AB|.

_{2}As |AB| is the tangent of |OP|, *m _{2}* is the negative reciprocal of

*m*. This means that we must fist find

_{1}*m*and then work our

_{1}*n*magic on it to get

^{-1}*m*.

_{2}**m _{1} = Δy/Δx**

*m*

_{1}= (y_{2}-y_{1})/(x_{2}-x_{1})*m*

_{1}= (4.415-0)/(2.347-0)*m*

_{1}= 4.415/2.347*m*

_{1}= (^{4415}⁄_{2347})Therefore:

*m _{2}* = -1/

*m*

_{1}*m*= -1/

_{2}*(*

^{4415}⁄_{2347})*m*= (

_{2}^{-2347}⁄

_{4415})

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 *m _{2}* of that line, we can use the equation

**to identify first point A and then point B. We start with A because A is the point where |AB| intersects the**

*y=mx+c**y*axis, and we know this to be

*c*in the equation of the line. Thus:

**y = mx+c**

*4.415 = ( ^{-2347}⁄_{4415})(2.347) + c*

*4.415 – 2.347(*

^{-2347}⁄_{4415}) = 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)/( ^{-2347}⁄_{4415}) = x*

*-5.663/(*

^{-2347}⁄_{4415}) = 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=√[(x _{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]**, where δ is |AB|. So let’s get that done now and wrap this baby up for the day:

**δ=√[(x _{2}-x_{1})^{2}+(y_{2}-y_{1})^{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.