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We have two non intersecting in $\mathbb R^2$ circles: $ C_1(x,y)\equiv x^2+y^2-2mx-2ny+p=0, $ $ C_2\equiv x^2+y^2-2m'x-2n'y+p'=0, $ where circles are with real coefficients and positive radiuses. Then they intersects in some $(a,b), (a',b')\in \mathbb C^2$. Let $C$ be an arbitrary circle, with or without real points, with real coefficients, passing throw $(a,b), (a',b')$. I wish to know, is $C$ is the form $\alpha C_1(x,y)+\beta C_2(x,y)$, for some real $\alpha, \beta$ ?

Thanks

Alex
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2 Answers2

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The answer is YES

Given $$C_1:x^2+y^2-2mx-2ny+p=0;\;C_2:x^2+y^2-2m'x-2n'y+p'=0$$ The linear combination

$$\mathcal{C}:\lambda C_1+\mu C_2=0$$ $\mathcal{C}$ gives the family of all circles, real or complex, passing through the intersection points of the two given circles.

For the special values $\lambda=1;\;\mu=-1$ we get the equation of the line joining the intersection points $(a,b), (a',b')$ wich is perpendicular to the line containing all the centers of $\mathcal{C}$

$(2m'+2m)x+(2n'-2n)y+p-p'=0$

Raffaele
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  • Maybe references to $$\mathcal{C}:\lambda C_1+\mu C_2=0$$ gives the family of all circles, real or complex, passing through the intersection points of the two given circles. – Alex Nov 14 '20 at 16:36
  • $\lambda C_1+\mu C_2=0$ passes through the intersection points because $C_1=0;;C_2=0$ implies $\lambda\cdot 0+\mu\cdot 0=0$ – Raffaele Nov 14 '20 at 16:40
  • Yes, but why if $C$ passes by the intersetion's points of two circles, then it has to be a linear combination of their equations. – Alex Nov 14 '20 at 16:47
  • because the intersection points satisfy both $C_1=0$ and $C_2=0$, thus any linear combination of the two – Raffaele Nov 14 '20 at 16:55
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The equation of a circle is

  • quadratic,

  • with equal coefficients of the squares, $x^2$ and $y^2$,

  • without mixed term $xy$.

Now the combination $\alpha C_1+\beta C_2$ is

  • quadratic,

  • with equal coefficients of the squares, $x^2$ and $y^2$,

  • without mixed term $xy$

hence it is the equation of a circle.

Plus, any point that belongs to both circles satisfies $\alpha C_1+\beta C_2=0$ and also belongs to that circle. So if the two given circles make two intersections, the whole family shares these intersections.

Notice that the coefficients of the squares can vanish, leaving the equation of a straight line as a degenerate case. For the same reason as below, this is the line through the two intersection points.

  • Thanks, but I don't know why $C$ passing by $(a,b), (a',b')$ is of the form $\alpha C_1+\beta C_2=0$. – Alex Nov 14 '20 at 16:24
  • @Alex: $(e^{C_1})^\alpha(e^{C_2})^\beta=1$ would also work. –  Nov 14 '20 at 16:27
  • I understand that $\alpha C_1+\beta C_2=0$ is a circle (or a complex straight line), and this circle contains $(a,b), (a',b')$ but I don't understand why $\alpha C_1+\beta C_2=0$ is equivant $C(x,y)=0$. – Alex Nov 14 '20 at 16:33
  • @Alex: $\alpha,0+\beta,0=0$. –  Nov 14 '20 at 16:49