Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Intersection
Replies: 1   Last Post: Aug 9, 2006 1:22 AM

 Search Thread: Advanced Search

 Messages: [ Previous | Next ]
 Rich Kleinschmidt Posts: 31 Registered: 12/6/04
Re: Intersection
Posted: Aug 9, 2006 1:22 AM
 Plain Text Reply

Your approach looks good. The algebra will be nothing that is beyond 2nd year algebra - just much uglier than any 2nd year text would ask you to tackle.

For example your 1st equation you get (y - b)^2 = Rab^2 - (x - a)^2
Then take the square root of both sides and get y = b +/- sqrt(Rab^2 - (x - a)^2)

Once you solve both equations for y, substitute for y and you get an equation two ugly radicals but only x. You can eliminate the radicals by isolating one and squaring both sides. Then isolating the other and squaring both sides.

You will need to keep track of the +/- ambiguities and possible extraneous solutions introduced by squaring. All the techniques are in 2nd year algebra.

Rich Kleinschmidt

-------------- Original message ----------------------
From: "Carlo Casini" <carlo_casini@worldnet.att.net>
> Hello,
> I need help on solving this problem.
> How do you find the intersection(s) of 2 circumferences knowing only the
> center and the radius of each one?
> I have look on several book and on the internet without any success.
> The only way that come to my mind is to represent the circumferences with a
> system of equation:
> (x - a)^2 + (y - b)^2 = Rab^2
> (x - c)^2 + (y - d)^2 = Rcd^2
> where a,b,c and d are the coordinate of the centers and Rab and Rcd the
> radiuses;
> I think x and y values should be the coordinate of the intersection(s)
> point(s).
> Any help would be greatly appreciated.

------- End of Forwarded Message

Date Subject Author
8/8/06 Carlo Casini
8/9/06 Rich Kleinschmidt

© The Math Forum at NCTM 1994-2018. All Rights Reserved.