Showing intersection multiplicity of circular points at infinity is two on the intersection of two concentric circles

24 Views Asked by At

This is in regards to a previous post that is still open here: Find mistake in calculations of intersection multiplicity to show that two circles have empty intersection

My question is, if we have two circles in $\mathbb{P}^2$, $\mathbb{V}(x^2 + y^2 - (z r_1)^2)$ and $\mathbb{V}(x^2 + y^2 - (z r_2)^2)$ with $r_1 \neq r_2$, how do we show that the intersection multiplicities of the two circular points at infinity on these circles is each two? My problem is that I keep getting that each is four instead of two.

Help Would be appreciated.