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This question was taken from Algebraic Geometry by Shafarevich.

1) Find a characterization in real terms of the line through intersection of two circles in the case that both of these points are complex.

2) Prove that it is the locus of points having the same power with respect to both circles.

What I have tried so far:

1) Take the two circles to be $ x^2+y^2+2f_1x+2g_1y+c_1=0 -----(1) \\ x^2+y^2+2f_2x+2g_2y+c_2=0 -----(2) $.

Take a point in $ \mathbb P^2, (\xi,\eta,\zeta). $

(1)-(2) $ \Rightarrow 2(f_1-f_2)x +2(g_1-g_2)y+c_1-c_2 = 0 $.

The point must satisfy this line. So,

$ 2(f_1-f_2)\xi+2(g_1-g_2)\eta+(c_1-c_2)\zeta=0. $

The points are complex when $ \zeta=0. $ So the equation of the line is $ (f_1-f_2)x+(g_1-g_2)y=0. $

Is this correct?

2) I need a hint for this part.

Thanks in advance for any replies.

pseudogeek
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1 Answers1

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No, this has absolutely nothing to do with projective coordinates.

If the equations of the real ($f_i,g_i,c_i$ are real numbers) circles are $x^2+y^2+2f_ix+2g_iy+c_i = 0$ then subtracting one from the other you get the equation $2(f_1-f_2)x + 2(g_1-g_2)y + (c_1-c_2) = 0$. Because it is a linear combination of the two equations, any point that is on both circles has to also be on that line.
Since circles only intersect at two points (in the affine plane), wether the coordinates of those points are complex or real, this is the equation of the real line joining them.

Geometrically, when the circles don't intersect (in the real affine plane) then this line is somewhere between the two circles (and it's also perpendicular to the line joining the two centers, for symmetry reasons). The two points of intersection are complex and you don't see them on the picture.

To prove that it is the locus of points that have the same power with regards to the circle, you have to write down the algebraic definition of power, then check that when you equate them you get the equation of a line, then check that the two points of intersection are on that line.

mercio
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