Chapter 9.2, Problem 46WE

To determine

To graph: Plot the graph for the given equation.

Explanation of Solution

Given information: It is given that $x=\sqrt{2y-y^2}$ and $\sqrt{a}\ge 0$

Calculation: As we know that a circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ with center $\left(h,k\right)$ and radius $\left(r\right)$ .

Now, the top half of the circle i.e. semicircle:

  $\begin{array}{l}x=\sqrt{2y-y^2}\\ x^2=2y-y^2\\ x^2+y^2-2x+1-1=0\\ x^2+\left(y^2-2x+1\right)=1\\ {\left(x-0\right)}^2+{\left(y-1\right)}^2=1^2\end{array}$

Compare ${\left(x-0\right)}^2+{\left(y-1\right)}^2=1^2$ to ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , we get,

• The center $\left(h,k\right)=\left(0,1\right)$
• The radius $r=1$

Graph: The required one $\left(x=\sqrt{2y-y^2}\right)$ is plotted below,