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I want to find the intersection coordinates of two functions:

  1. A quadratic function $f$ of this style: $f(x) = a(x-b)^2+c$
  2. And a circle function $g$: $(x-d)^2+(y-e)^2=r^2$ with $y=g(x)$ and $r$ being the radius of the circle.

I know that I can rewrite the second function to $g(x)=\sqrt{r^2-(x-d)^2}+e$

Now I find myself in the position to solve $$ \sqrt{r^2-(x-d)^2}+e =a(x-b)^2+c $$ towards $x$ since I should find there zero to two intersections. But how can I find them? If I square both sides I will have to deal with $x^4$ - any ideas?

Olaf007
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  • Bear in mind that the radical needs a $\pm$, since at present it only gives the upper half of the circle. (Each half is a function of $x$, but not the entire circle, since a function cannot send one thing to two different values.) – PrincessEev Jul 20 '22 at 08:25
  • Does $e$ stands for exponential or just constant $e$? – KMN Jul 20 '22 at 08:28
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    It might help if you start with a change of coordinates so that either the parabola or the circle is a simpler equation - like $y=x^2$ or $x^2+y^2=1$. – Peter Jul 20 '22 at 08:36
  • There can be up to four intersections between a parabola and a circle, so I would expect to get a quartic equation. – Jaap Scherphuis Jul 20 '22 at 09:05
  • You don't need to use square roots too early, otherwise you will need square it back again. Equation $(x-d)^2+(a(x-b)^2+c-e)^2-r^2=0$ is algebraic equation of fourth degree. There is general method for solving quartic equations, but general solution is too long to write it here. Quartic equation may have 0,1,2,3,4 distinct real solutions, which are corresponding to 0,1,2,3,4 intersection points. There cannot be two intersection points with the same $x$, because every $x$ has only one corresponding $y=a(x-b)^2+c$. – Ivan Kaznacheyeu Jul 20 '22 at 10:01

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Since $f(x)=y$ just plug the value of the quadratic function in the equation of the circle. There could be from zero to $4$ intersections i.e $4$ zeros, not only $2$ . Bezout’s theorem states that a curve of degree m and a curve of degree n intersect at most $mn$ times. Applying this to a circle (degree 2) and a parabola (degree $2$) we get at most $4$ intersections.

Guillermo García Sáez
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