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**jimi70****Member**- Registered: 2012-04-04
- Posts: 22

'The circle S1 with centre C1(a1, b1) and radius r1 touches externally the circle S2 with centre C2(a2, b2) and radius r2. The tangent at their common point passes through the origin. Show that

(a1² - a2²) + (b1² - b2²) = (r1² - r2²).

If, also, the other two tangents from the origin to S1 and S2 are perpendicular, prove that

|a2b1 - a1b2| = |a1a2 + b1b2|.'

I can do the first part but need help with the second part.

Thanks for your time.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,337

hi jimi70

I expect you used Pythagoras on those right angled triangles to do the first part.

It looks to me like you can do it again a few more times.

The new tangents will be equal in length to the first. I've shown all three in red in my diagram.

I haven't tried it yet but I think it'll work out. But there'll be a lot more algebra to do. Post back please; either to say it worked or to ask me to have a proper go at it myself.

EDIT: The line from C1 to C2 goes through the tangent touching point:

That may help.

FURTHER EDIT: I haven't done it yet but I'm still trying ......................

Bob

*Last edited by bob bundy (2013-11-25 02:27:47)*

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,337

getting nowhere so I thought I'd try some values.

See revised diagram below.

I'm not getting the required result. ???

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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jimi70 wrote:

'The circle S1 with centre C1(a1, b1) and radius r1 touches externally the circle S2 with centre C2(a2, b2) and radius r2. The tangent at their common point passes through the origin. Show that

(a1² - a2²) + (b1² - b2²) = (r1² - r2²).

Let be the point of contact of the circles.

The tangent through this point is perpendicular to the line joining their circles, which has gradient

; therefore the tangent through the common point has gradient . In other words, the point lies on the line :The equations of the circles are

and ; since lies on both of themNow subtract [3] from [2] and use the relation [1], and you should get the answer.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,337

Nehushtan wrote:

and you should get the answer.

Which one?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

He's probably asking because the OP has stated that it is the second part he is not able to do.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,337

hi Nehushtan

Nehushtan wrote:

Why dont you read my proof?

I did and it left me very confused.

Post 1 has two parts and jimi70 says he can do part 1. I looked at it and did it in my head. If I wrote it out, it would take 3 lines.

Part 2 has extra information. I couldn't see where you had used that information.

I had already spent some time trying to prove part 2 and failed. I made an accurate diagram of what I think is the correct situation (with particular coordinates) and the required result did not come out. Now, it's entirely possible that I have my diagram wrong. Your opinion on that would be helpful.

So then I looked at your answer. I subtracted as you suggest and used (1) to eliminate A and B but I still had the radii to deal with. So I used the equations from part 1 to eliminate them and arrived back at your equation 1.

I'm fairly certain that part 2 won't come out without using the additional information to make an extra constraint, so it doesn't surprise me that I went round in circles. If I've got it wrong, please supply the missing steps in your proof. I'd be most grateful for another view on this, thanks.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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