Show $y= \sqrt{2x-x^2}$ maps out a circle of radius centered at the coordinate $(x,y) = (1,0)$

Question

Show $y= \sqrt{2x-x^2}$ maps out a circle of radius centered at the coordinate $(x,y) = (1,0)$. Further show that the circle is described by the equation $r = 2 \cos \phi$ ($\phi$ is defined to the x axis)

How am I suppose to approach this? I am aware that y and x have to a positive value $\sqrt{2x-x^2}$, but what does $r = 2 \cos \phi$ has to do with this?

Answer 1

First consider that

\begin{equation*} y = \sqrt{2x-x^{2}} \implies y^{2} = 2x-x^{2} \end{equation*}

and, completing the square for $2x-x^{2}$, we obtain

\begin{equation*} y^{2}+(x-1)^{2} = 1 \end{equation*}

which is a circle of radius $1$ centered at $(1,0)$. Actually, though, $y = \sqrt{2x-x^{2}}$ only maps out the top half of a circle of radius $1$. You need $y = \pm\sqrt{2x-x^{2}}$ to get the whole thing.

After that, recall that $(x,y)$ in the $xy$-plane can be written as

\begin{equation*} (r\cos{(\phi)},r\sin{(\phi)}) \end{equation*}

for some $r>0$ and $0\leq \phi < 2\pi$. So you have that

\begin{equation*} y^{2} + (x-1)^{2} = r^{2}\sin^{2}{(\phi)} + r^{2}\cos^{2}{(\phi)}-2r\cos{(\phi)}+1 = 1 \end{equation*}

whereby

\begin{equation*} r^{2} - 2r\cos{(\phi)} = 0\implies r = 2\cos{(\phi)}. \end{equation*}

Answer 2

If $x,y\in\mathbb{R}$ are such that $y=\sqrt{2x-x^2}$, then $y^2=2x-x^2$. In other words, $(x-1)^2+y^2=1$. So, $(x,y)$ belongs to the circle centered at $(1,0)$ and radius $1$.

And if you take $(x,y)$ from that circle and you write $x$ as $r\cos\phi$ and $y$ as $r\sin\phi$, with $r>0$, then \begin{align} y^2=2x-x^2&\iff r^2\sin^2\phi=2r\cos\phi-r^2\cos^2\phi\\ &\iff r^2=2r\cos\phi\\ &\iff r=2\cos\phi. \end{align}

Answer 3

A bit generalized, and also a bit overfit solutions to this...

$y=\sqrt{2ax-x^2}$ maps out a circle of radius $a$ centered at the coordinate $(x, y)=(a, 0).$ Also, the circle is described by the equation $r=2a\cos{\phi}$.

Part I. $y=\sqrt{2ax-x^2}$ maps out a circle of radius $a$ centered at the coordinate $(x, y)=(a, 0)$.

$y=\sqrt{2ax-x^2} \Leftrightarrow (x-a)^2+y^2=a^2$, proved.

Part II. The circle is described by the equation $r=2a\cos{\phi}$.

\begin{align} & \text{let } y=r\sin{\phi}, x=r\cos{\phi}. \\ & \Rightarrow \; r^2\sin^2\phi=2ar\cos\phi-r^2\cos^2\phi. \\ & \Rightarrow \; r^2=2ar\cos\phi, r=2a\cos\phi. \end{align}