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I hope you're well,

I was finding the points of intersection of $$ x^2 + (y-1)^2 = 1 \quad \text{and} \quad y = 1-x^2 $$ If I rearrange the formula of the circle to $$ (y-1)^2 = 1-x^2 $$ then substitute the second equation to get $$ (y-1)^2 = y $$ I then solve this to get two values of $y$, of which obviously by the symmetry of the problem there can only be one value of $y$. Where has this additional 'solution' come from? Is it because I substituted something that was a quadratic with something that was linear?

Thanks!

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  • $\begingroup$ You have to enclose the MathJax in $ signs for the formatting commands to take effect. $\endgroup$ – saulspatz May 15 at 19:41
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There is more going on here than meets the eye, and I mean it.

You have two conics, i.e. curves of degree two, and according to Bézout’s Theorem on Curves, there are four intersection points, when properly counted. In this case, proper counting means finding the points with complex coordinates as well. Run through the computation, and you see that these are $$\left(\pm\sqrt{\frac{-1-\sqrt5}2},\frac{3+\sqrt5\,}2\right)\,.$$ Since the real eye sees only the real points, which I’m sure you found, that’s the real story, as far as it goes. But the mind’s eye sees the complex points as well, and there they are.

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Well, if $\color{red}{y}={\color{green}{1-x^2}}$ and $$x^2+(\color{red}{y}-1)^2=1$$ Then:

$$x^2+([\color{green}{1-x^2}]-1)^2=1$$ $$\to x^4+x^2-1=0$$

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It is because $(y-1)^2=y$ does not imply that there exists a real $x$ such that $y=1-x^2$ (which can only be $\le 1$). In oher words, the set of solutions to "$A=B$ and $A=C$" may be smaller than the sret of solutions to "$B=C$". In fact, in your case "$(y-1)^2=y$" has way more solutions than the original system, namely two values (instead of one) of $y$ and for each infinitely many (instead of two) of the no longer involved $x$. I'd say the latter is even more dramaitic than the former...

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