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Proving the Equation of the Pencil of Two Intersecting Circles

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Given two circles algebraically represented by the equation

$$ x^2 + y^2 + 2g_{1}x + 2f_{1}y + c_{1} = 0 $$ $$ x^2 + y^2 + 2g_{2}x + 2f_{2}y + c_{2} = 0 $$

in this post I will be talking about the pencil of circles passing through the points of intersection of the above circles. For the sake of simplicity here, let's assume that the two circles are indeed intersecting at two separate points.

I will be representing the two equations above using

$$ C_{1} = 0 $$ $$ C_{2} = 0 $$

to be concise.

I am trying to prove that the pencil of circles here is

$$\lambda C_{1} + \mu C_{2} = 0$$

where $\mu$ and $\lambda$ are arbitrary constants. Wikipedia does have a proof for this (in fact it is for all conics) but it went straight over my head. I'm currently in high school and I don't know enough to understand those kind of proofs.

However, I did come up with a proof on my own, which I'm a bit skeptical about since it seems "too easy". It is:

When we add the two equations $C_{1} = 0$ and $C_{2} = 0$ after multiplying each of them with any arbitrary constant, we get an equation which still represents a circle (since the coefficient of $x^2$ and $y^2$ are equal to each other, and the coefficient of the $xy$ term in the general conic equation we get is zero). Now we simply need to prove that this new resulting circle, that is,

$$ \lambda C_{1} + \mu C_{2} = 0$$

passes through the original intersection points.

The points of intersection are on both the circles and hence satisfy $C_{1} = 0$ and $C_{2} = 0$. Therefore the above equation of the new resulting circle satisfies the points of intersection.

QED.

Is this proof correct? If it is, can it be extended to other conics? (the proof will remain pretty much the same I'm guessing)

Could you provide other simplistic proofs for this?

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1 answer

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The proof is not quite correct. I mean, it's correct as far as it goes, but all it proves is that the curves $\lambda C_1+\mu C_2=0$ are contained in the desired pencil. You haven't proven that every circle that goes through your two points is of the form $\lambda C_1+\mu C_2=0$.

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3 comments

ohh, how did I miss that. I'll try proving that by myself later, will edit the post if I come up with a proof. TripleFault‭ about 1 month ago

@TripleFault I don't think it would be appropriate to edit your question post to include that proof. After all, you asked whether your proof was good, I gave a good answer. If you now edit your question to make my answer nonsensical, that's unfair to me (making me look foolish) and to readers (making the page as a whole unclear to them). Better is to leave your question here as is and, if you have a new question, to ask it separately. msh210‭ about 1 month ago

Okay, I won't, but I don't see why it would be such a bad idea to do so. I'll add in the proof at the end of the post with a clear demarcation "edit", and your answer would still be useful to viewers since it tells any viewer why that extra part was needed in the first place; I feel it actually might benefit viewers to show the part which was missing, since that not only tells them why my proof was wrong but also what needs to be added to correct it. Still, I won't be editing the post, thank you TripleFault‭ about 1 month ago

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