Equation of a constrained circle

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Sven
Sven 2013 年 5 月 18 日
I have a circle with known parameters (x,y,r):
x = 37
y = -7
r = 38
I would like to find the parameters of a new circle (x2,y2,r2) that:
  1. Passes through the origin [0,0]
  2. Passes through the point [x,y+r]
  3. Keeps the same x value (ie, x2==x)
Can anyone wrap their head around this one? Basically it's like pinning the max-y point on the first circle, and then growing-shrinking that circle until it crosses the origin.
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Roger Stafford
Roger Stafford 2013 年 5 月 18 日
編集済み: Roger Stafford 2013 年 5 月 18 日
It's better if you solve it yourself (with a little help.) Let (uppercase) X and Y be arbitrary coordinates on your new circle with unknown parameters x2, y2, and r2. Now write the above three conditions as the three equations these quantities must satisfy and solve for x2, y2, and r2. I'll get you started on the first equation:
(0-x2)^2+(0-y2)^2 = r2^2 <-- substituting (0,0) for (X,Y)
Sven
Sven 2013 年 5 月 18 日
Thanks Roger. Don't worry, not homework... I was just about to get on a long flight and this was bugging me. With your prompting I got there. I'd gone down that route but used wolfram alpha without taking the time to think about it... it doesn't like variables called x2 (interprets them as 2*x). I got flustered and was in a hurry, hence the question here :)
As a W.A. string: (0-x)^2+(0-b)^2 = c^2, (y+r-b)^2 = c^2, solve for b, c

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回答 (1 件)

Youssef  Khmou
Youssef Khmou 2013 年 5 月 18 日
hi try to verify this initiation :
the first circle is defined by :
a=37;b=-7;r=38;
We are looking for new circle based on The three conditions :
1.a2^+b2^=r2^2
2.(a-a2)²+((b+r)-b2)^2=r2^2
3.a2=a;
the solution is ( to be verified ):
a2=a;
b2==((b+r)^2+a^2)/(2*(b+r));
r2=r2=b+r-b2;
....

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