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Let me ask a similar question to the one I did yesterday. I got answers which said that the following problem had no general solution for x and y.

$\sqrt{(n_1-x)^2+(n_2-y)^2}=n_3$

$\sqrt{(n_4-x)^2+(n_5-y)^2}=n_6$

How about searching in complex numbers? Does it have the general solution in the set of complex numbers?

Thank you in advance.

In details, all in $\lbrace n_1,...,n_6\rbrace$ are real numbers. And I want a formula for the complex number solutions for (x,y) . You can forget the word "circle" in the title .

Previous Question

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Are you asking whether two circles have (complex) points in common? Any assumptions about $n_1,\dots,n_6$? –  Bill Cook Nov 9 '11 at 19:27
1  
Let the $n_i$ be real, with $n_3, n_6\ge 0$. There is an easy way to look for the real solutions. Square both sides of each equation, subtract. We get a linear equation. Substitute. We get a quadratic in one variable, easy to solve in general. Solution needs not be real since the circles need not meet. But the procedure gets us all the solutions (which may be non-real) always. –  André Nicolas Nov 9 '11 at 19:35
    
@BillCook,I edited the question to meet your comment.Thank you very much. –  seven_swodniw Nov 10 '11 at 4:50
    
@AndréNicolas,won't your procedure be a formula?Thank you very much. –  seven_swodniw Nov 10 '11 at 4:52
    
It will yield a formula, sure. It is essentially the same as the one by Swapan. –  André Nicolas Nov 10 '11 at 5:26

1 Answer 1

up vote 1 down vote accepted

As I answered your previous question (thanks for your acceptance), I feel I should clarify things. So let me try at my best.

First of all it looks (from your previous question) you are trying to get some sort of program which would take the following inputs: $x_1,x_2,y_1y_2,d_1,d_2$ and output the point $p$ if it exists (or some variant of this). I will try to put the algorithm at the end of this answer. Lets first clarify when there will be a solution and when not.

So, we have two equations in hand $$(x-x_1)^2+(y-y_1)^2=d_1^2\quad\cdots(1)$$ $$(x-x_2)^2+(y-y_2)^2=d_2^2\quad\cdots(2)$$ Subtracting (2) from (1), we get a linear equation of the form $$ax+by=c\quad\cdots(3)$$ where $$a=2(x_2-x_1),b=2(y_2-y_1),c=d_1^2-d_2^2-x_1^2-y_1^2+x_2^2+y_2^2\quad\cdots(4)$$

Note that for distinct points, at least one of $a,b$ is non zero, WLOG, say $b$. Now substituting the value of $y(=\frac{c-ax}{b})$ into (1), we get a quadratic equation of the form $$Ax^2+Bx+C=0\quad\cdots(5)$$ where $$A=a^2+b,B=2(aby_1-ac-bx_1), C=bx_1^2+(by_1-c)^2-d_1^2b^2\quad\cdots(6)$$ We are almost done now. Note that the equation (5) has a (REAL) solution (which would imply the point $p$ exists) iff either $A=0$, Or $(A,B)\ne(0,0)$ and $$D=B^2-4AC\ge0\quad\cdots(7)$$ So, assuming$A^2+B^2\ne 0$, there exists no $p$ if $D<0$, a unique $p$ if $D=0$ and two $p$s if $D>0$.


A very rough algorithm: (My apology, I don't know much about these)

Input $x_i,y_i,d_i$

Define $a,b,c$ as given in (4)

If $b\ne0$

Define $A,B,C,D$ as given in (6) and (7)

Else Define $A,B,C,D$ assuming $a\ne0$ in (3) *\Note that we use the same variable $A,B,C$

If ($A^2+B^2=0$, or $D<0$): No $p$ exists

Else If $A=0$: Unique $p$ *\The expression for $p$ may change on the assumption $a,b\ne0$ in (3), though

Else If $D=0$: Unique $p$

Else: Two distinct $p$.


Final comments: 1. I hope now you can do it far better than me. 2. There may lots of typo, I do them very often. 3. Anyone wish to improve, change, edit this answer is warmly welcome.

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"Subtracting (2) from (1), we get a linear equation..." - you can say you've taken the radical line of the two circles. –  J. M. Nov 9 '11 at 23:03
    
@Sapan , thank you for the answer very much. It is very helpful. –  seven_swodniw Nov 10 '11 at 4:30
    
@J.M.: Now I see what this "radical line" is. Honestly I missed your comment in the other answer of previous question :) –  Tapu Nov 10 '11 at 8:36

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