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Determine the equation of the circle reflection of the circle $x^2 + y^2 = 1$ if the circle of reflection is $x^2 + y^2 + 2x = 0$. I'm learning about circle inversion but I still don't get what this question is saying? Is it saying find an equation that reflects $x^2+y^2=1$ to $x^2 + y^2 + 2x = 0$?

user60887
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    No, reflect the unit circle in the other circle (to get another circle). –  Apr 30 '13 at 02:41
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    so I graphed it on wolfram alpha and both circles intersect at (-0.5,0.866025) and (-0.5,-0.866025). Oh wait im trying to find the equation that's going to be of the form $ax+by+c=0$. I think I get it. – user60887 Apr 30 '13 at 03:03
  • No,it is saying: find an equation of the reflected circle that reflects this given circle $x^2 + y^2 + 2x = 0$? about a unit mirror circle of inversion $x^2+y^2=1$. – Narasimham Sep 06 '14 at 10:18

2 Answers2

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You have two circles. The second one defines your transformation: it is an inversion in that second circle. The first one is the object to which you want to apply that transformation, resulting in a new object for which you should find the equation. Here is an illustration:

Image of configuration

The black circle is $x^2+y^2+2x=0$. The blue circle is $x^2+y^2=1$. Since the blue circle passes through the center of the black circle, a circle inversion will map that central point to the point at infinity (of $\hat{\mathbb C}$). So the resulting object will not be a circle but instead a line (a circle with infinite radius, if you want).

Hence the blue line in the image above is the result of applying an inversion in the black circle to the blue circle.

Finding its equation should be easy from the image, but in order to understand this task, you might want to think about how you'd find that equation without an image, or in cases where deducing the numbers from the image would be less easy.

MvG
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The question as given here on the topic of pure inversion is confused. Reflection and Inversion are synonymous here. Better to ask the question as:

Determine the equation of the reflected circle from given circle $x^2 + y^2 + 2x = 0$ when the reflecting circle or circle of inversion is the unit circle $x^2 + y^2 = 1.$

Answer is: x = -1/2.

You can use my code in Mathematica for practice to sketch circles before and after reflections about the Red circle/mirror of inversion.

h = 5.5; R = 3 ; {x[t_], y[t_]} = {h + R Cos[t], R Sin[t]} ;

c = ParametricPlot[{x[t], y[t]} , {t, .001, Pi - .001}];

Rinv = 4;{x1[t_], y1[t_]} = Rinv^2 {x[t], y[t]}/(x[t]^2 + y[t]^2);

c1 = ParametricPlot[{x1[t], y1[t]} , {t, .001, Pi - .001}];

cINV = ParametricPlot[Rinv { Cos[t], Sin[t]} , {t, .001, Pi - .001}, PlotStyle -> {Red, Thick}];

Show[{c, c1, cINV}, PlotRange -> All]

Narasimham
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