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](https://en.wikipedia.org/wiki/Pencil_(mathematics)) 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)](https://en.wikipedia.org/wiki/Pencil_(mathematics)#Pencil_of_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](https://en.wikipedia.org/wiki/Conic_section#General_Cartesian_form) 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?
TripleFault
·
2020-10-29T12:10:19Z (about 4 years ago)