Re: Finding origin of circle

From: Hans van Duijnhoven (hanxs.gerda_at_mail.tele.dk)
Date: 06/22/04 

Date: Tue, 22 Jun 2004 23:08:42 +0200

On Tue, 22 Jun 2004 20:47:50 +0200, "Johan Karl Larsen"
<jklnews1@itlarsen.net> wrote:

>> There are two possible origins. Given point (x1, y1), the orgin must
>> lie on the circle (x-x1)^2 + (y-y1)^2 = r^2. Similarly, (x2, y2) gives
>> the origin on the circle (x-x2)^2 + (y-y2)^2 = r^2. The intersection of
>> these two circles gives you the possible locations of the center.
>>
>
>Yes, and how do I solve two quadratic equations with two unknowns? Easy if
>it was linear by using elimination.
>
>Once I have the two possible solutions I can reject the one that is in the
>wrong quadrant.

You have circles: (x-x1)^2 + (y-y1)^2 = r^2
and (x-x2)^2 + (y-y2)^2 = r^2
If you subtract both equations, you have
(x-x1)^2 -(x-x2)^2 + (y-y1)^2-(y-y2)^2 = 0
(x-x1+x-x2)*(x-x1-x+x2) + (y-y1+y-y1)*(y-y1-y+y2)=0
(2x-x1-x2)*(x2-x1) + (2y-y1-y2)*(y2-y1)=0
which is linear in x and y.
Now you can write x= .... (or y=...)
and put the result in one of the circles to solve for y (or for y)

Hans van Duijnhoven,
Randers, Denmark.

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