Use the several variable version of the Newton-Raphson algorithm to find all intersection points (real and complex) of the 2 circles $$(x-7)^2+(y-2)^2=100, \text{ and }$$ $$(x-11)^2+(y-5)^2=75$$

So I was able to find the real solution easily by plotting both circles and determining where they both intersect and using that estimate to use the Newton-Raphson algorithm.

However, finding the complex solutions have been difficult. I know I can just guess a point with complex values and use the Newton Raphson method. But all my guesses don't work out.

After asking my Professor for a method to efficiently find a guessing point. He suggested to use the modulus and note that

$$\omega=|f(x,y)|+|g(x,y)|=0$$ is true if and only if both $f(x,y)=0$ and $g(x,y)=0$, where $$f(x,y)=(x-7)^2+(y-2)^2-100, \text{ and }$$ $$g(x,y)=(x-11)^2+(y-5)^2-75$$

So suppose that $(a+b*I,c+d*I)$ is a solution to both $f$ and $g$, but then I have two equations with four unknowns.

So can I have some at advice how to get those complex solutions?

  • $\begingroup$ This might help mathpages.com/home/kmath689/kmath689.htm It will certainly give you two extra equations so that you have four equations in four unknowns. $\endgroup$ – Carl Christian Feb 23 '16 at 20:53
  • $\begingroup$ Should there even be any complex solutions? Given the equations of two circles we can take a linear combination that cancels out all the quadratic terms. That deflates the equation for x or y alone from a quartic to a quadratic leaving only two roots, which in this case are the real ones. $\endgroup$ – Oscar Lanzi Feb 24 '16 at 11:16

You are correct; set $x=a+i b$, $y=c+id$, replace in the equations and isolate the real and imaginary parts (setting them equal to zero).

Doing so, you get $$a^2-14 a-b^2+c^2-4 c-d^2-47=0 \tag 1$$ $$2 a b-14 b+2 c d-4 d=0\tag 2$$ $$a^2-22 a-b^2+c^2-10 c-d^2+71=0\tag 3$$ $$2 a b-22 b+2 c d-10 d=0\tag 4$$ You could solve these four equations using Newton-Raphson method.

What I found amazing is that the problem has analytical solutions.

$(1)-(3)$ leads to $$8 a+6 c-118=0 \tag 5$$ Similarly $(2)-(4)$ leads to $$8 b+6 d=0\tag 6$$ So, using $c=\frac{59-4 a}{3} $ and $d=-\frac{4 b}{3}$ and replacing we are let with $$a^2-22 a-b^2+94=0\tag 7$$ $$a b-11 b=0\tag 8$$ The last equation shows that either $a=11$ or $b=0$.

If $a=11$, $(7)$ gives $b^2+27=0$ that is to say $b=\pm 3 i \sqrt{3}$

If $b=0$, $(7)$ gives $a^2-22 a+94$ that is to say $a=11 \pm 3 \sqrt{3}$ For each pair, compute the corresponding $c$ and $d$ and the four solutions are $$\left\{a= 11,b= -3 i \sqrt{3},c= 5,d= 4 i \sqrt{3}\right\}$$ $$\left\{a= 11,b= 3 i \sqrt{3},c= 5,d= -4 i \sqrt{3}\right\}$$ $$\left\{a= 11+3 \sqrt{3},b= 0,c= 5-4 \sqrt{3},d= 0\right\}$$ $$\left\{a= 11-3 \sqrt{3},b= 0,c= 5+4 \sqrt{3},d= 0\right\}$$

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    $\begingroup$ The complex solutions for a, b, c, d given above just reduce to the same x and y values as the real roots. There are, in fact, no complex conjugate roots! $\endgroup$ – Oscar Lanzi Feb 24 '16 at 11:19
  • $\begingroup$ Okie dokie. Now I agree with what everyone was saying about there not being any complex conjugate roots. But I needed to see it through some kind of work to believe it. How did we get $d=-4b/3$? $\endgroup$ – Username Unknown Feb 24 '16 at 21:20
  • $\begingroup$ @UsernameUnknown $d=−4b/3$ comes straight from equation $(6)$ $\endgroup$ – AakashM Feb 26 '16 at 8:26

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