Turning point of equation

Question

Using differentiation, find the turning points of $$ (x^{2}+y^{2}-x)^2=x^{2}+y^{2} $$

Thanks!

Answer 1

We have \begin{equation} (x^{2}+y^{2}-x)^2=x^{2}+y^{2}\tag1\label{eq:1} \end{equation} Differentiate both sides and put $y'=0$.

\begin{align*} &(x^{2}+y^{2}-x)^2=x^{2}+y^{2}\\ \implies &2(x^2+y^2-x)(2x+2yy'-1)=2x+2yy'\\ \implies &(x^2+y^2-x)(2x-1)=x\tag2\label{eq:2} \end{align*} Using \eqref{eq:2} in \eqref{eq:1}, we get, \begin{align*} \left(\dfrac{x}{2x-1}\right)^2=x^2+y^2\tag3\label{eq:3} \end{align*} Using \eqref{eq:3} back into \eqref{eq:2}, we get, \begin{align*} &\left(\dfrac{x}{2x-1}\right)^2-x=\dfrac{x}{2x-1}\\ \implies &\left(\dfrac x{2x-1}-\dfrac12\right)^2=x+\dfrac14\\ \implies &\dfrac{1}{4(2x-1)^2}=\dfrac{4x+1}{4}\\ \implies &(4x+1)(2x-1)^2=1\tag4\label{eq:4} \end{align*} Now, \eqref{eq:4} is a cubic equation with $0$ as one of the roots.