4 intersecton points of 2 conic circles are always concyclic?

Question

Given 2 conic section of type

$$ax^2+2hxy+by^2+2gx+2fy+c=0$$ have 4 intersection points. Are the points for sure concyclic?

If yes, does the proof come from the fact that I can create an infite amount of new conic section using the linear combination of them and there is for sure a value of lambda that will give me a circle?

Welcome to MSE! Please use the basic tutorial and quick reference guide and also show the work you have done so far. — Jessie, Feb 03 '21 at 13:29

Jan-Magnus Økland · Answer 1 · 2021-02-03T15:05:07.137

1

No. In geogebra varying coefficients you can easily find counterexamples.

This has solutions

$(1,0),(\frac1{\sqrt{3}},\frac1{\sqrt{3}}),(-\frac1{\sqrt{3}},-\frac1{\sqrt{3}}),(\frac37,-\frac87).$

Which you can check are not concyclic: $(x-1/6)^2+(y+1/6)^2=13/18$ does not contain $(\frac37,-\frac87).$

edited Feb 03 '21 at 15:05

answered Feb 03 '21 at 13:51

Jan-Magnus Økland

5,149

how would i find the coefficients of the polynomials to disprove this? – Nethanel Benzak Feb 03 '21 at 14:30
or how can u disprove this in a formal way? – Nethanel Benzak Feb 03 '21 at 14:31
@NethanelBenzak: See edit. – Jan-Magnus Økland Feb 03 '21 at 15:05
This is a question on an exam. How did u calculate the conic section and know which circle to fit and know which points would disprove this? – Nethanel Benzak Feb 04 '21 at 16:15

4 intersecton points of 2 conic circles are always concyclic?

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