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How do we find the degree of this equation: $$ \sqrt{\left(x-a_{1}\right)^{2}+\left(y-b_{1}\right)^{2}}+\sqrt{\left(x-a_{2}\right)^{2}+\left(y-b_{2}\right)^{2}}+\sqrt{\left(x-a_{3}\right)^{2}+\left(y-b_{3}\right)^{2}}+\sqrt{\left(x-a_{4}\right)^{2}+\left(y-b_{4}\right)^{2}}=R $$

Where $(a_1,b_1),(a_2,b_2),(a_3,b_3),(a_4,b_4)$ are 4 distinct points in the plane, and $R$ is a constant.

Jeeth2006
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  • I don’t know if it answers your question, but [this post](https://math.stackexchange.com/questions/124333/what-are-curves-generalized-ellipses-with-more-than-two-focal-points-called-an) may help – Gauss Jan 07 '23 at 15:50
  • Thanks i hadn't come across that post before but i am still curious how they rationalised √a+√b+√c+√d = k – Jeeth2006 Jan 07 '23 at 16:02
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    There's more details in here [k -Ellipse](https://doi.org/10.48550/arXiv.math/0702005). – koy Jan 07 '23 at 16:08
  • Another link : https://math.stackexchange.com/a/3866928/305862 – Jean Marie Jan 07 '23 at 16:24
  • @Gauss Wouldnt that give a+b+2√ab=k^2+c+d-2k(√c+√d)+2√cd, so we still have 4 roots – Jeeth2006 Jan 07 '23 at 16:26
  • @Jeeth2006 You’re right, sorry for the mistake – Gauss Jan 07 '23 at 18:17

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