Tangents of a circle passing through a point
An example of this:
(you can ignore the dotted circle)
My friend and I had difficulty trying to solve it at first. Generating a line equation is easy, but finding the gradient isn't.
We spent 3 hours over 3 days at school (we have daily 1 hour long "Study" classes, which most people don't "study" in anyway), and this is pretty much how it ended up:
- 1st day: I solved the question by using a multitude of trigonometric functions, ending up with the two gradients. But they were not in any particular form, just numbers, so weren't that great as an answer as they would not be exact answers. And it looked ugly (yes, maths can be beautiful at times...)
- 2nd day: So my friend attempted to get a surd form for this number by working through my method, and he successfully does.
- 3rd day: He wanders off and discovers a relationship between the answer and original parameters and generates a general form for the gradient for the lines!
So today I copied off all our rough work onto paper and our teacher was most pleasantly surprised!
The equation doesn't involve having to substitute one equation into another, which has significantly made calculating the gradient for any line of this sort a whole lot easier.
But I won't be posting it up. Not yet.
That's because I'm trying to find if this has been generalised before. So if anyone knows of anything which can directly generate the gradients of such lines, please let me know in the Comments section below.
Otherwise... well, you can probably guess what we will do next.
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Comment by bazzer, on 12-Mar-2010 10:54 , user id: 31191)
My first thought was you could just differentiate the equation of the circle to get the gradient at a point. Since the lines are tangential, you know the relationship between their gradients, so it's easy? But then, I haven't tried to do it yet ;)
Comment by bazzer, on 12-Mar-2010 10:56 , user id: 31191)
Actually, you don't need the gradient of the radius line do you, so it's even easier... isn't it?
Comment by bazzer, on 12-Mar-2010 11:15 , user id: 31191)
Or thinking about it, isn't it super easy to calculate the gradient of the radius? Then work out the gradient of the tangent from that? i.e. the tangent is at (a,b) where a^2+b^2=r^2, i.e. (a, (r^2-a^2)^0.5). So the gradient of the radius is (r^2-a^2)^0.5/a. The gradient of the tangent is therefore -a/(r^2-a^2)^0.5 and we know the point it travels through so the equation is:
y = (-ax + r^2)/(r^2-a^2)^0.5
Well, I haven't checked it, so it's probably wrong somewhere.
Comment by bazzer, on 16-Mar-2010 13:29 , user id: 31191)
I guess I didn't understand your question then. Are you given the circle and the point and you want to generate the equations of the lines? Or are you given the point and you want to generate the equations of the lines and an arbitrary circle that they form tangents with?
Obviously in my third comment a and b are not given, but can be calculated. I did assume the centre of the circle was at the origin (O) but this can easily be transformed to any location.
Safe to say, if you and your friend could solve this in 3 days then one of the following applies (in order of likeliness):
a) It's been solved before
b) You got something wrong
c) You're genii
I'd really like the exact wording of the question, so I can have a proper go at solving it. I feel that your blog post above is a bit vague (or maybe I'm a bit dense).
And I don't want to be mean, but by the sounds of it you're not quite a mathematician yet.
Author's note by manhinli, on 16-Mar-2010 17:35 , user id: 31860)
@bazzer - Basically we're given a point which is some distance away from the circle, and told to find two lines which pass through the said point which are tangents to the circle.
By order of likeness, obviously it's been solved before, and we didn't make something wrong. And most definitely not genii since we're still students who aren't as spectacular as the one way up in the sky...
But meh. When I put up our little thing up (some time in the near future) it'll look dumb and short (like "What? That took you three days?" 
) - but sometimes maths is like a journey into a wild world of numbers...
Comment by michael leekens, on 6-May-2010 11:17 , user id: )
Could you please post the solution??
I have been waiting for a while now and haven't figured it out..
cheers
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Comment by nzwaz, on 11-Mar-2010 23:08 , user id: 33733)
http://mathworld.wolfram.com/CircleTangentLine.html
Google it ??