Chapter 8, Problem 16REAE

Solution

Graph the given equation and identify the radius of the circle.

The graph of the given equation is

  

Thus, the radius of circle is $2\sqrt{10}$ units with centre $\left(0,0\right)$ .

Given information:

Given equation is:

  $x^2=40-y^2$

Formula used:

Standard form of equation of a circle with radius ‘ r’ and center $\left(h,k\right)$ is

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation:

The given equation is

  $x^2=40-y^2$

Or

  $x^2+y^2=40$

In order to graph the given equation, first rewrite the equation as,

  $\begin{array}{l}{\left(x-0\right)}^2+{\left(y-0\right)}^2={40}^2\\ {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(\sqrt{40}\right)}^2\end{array}$ ,

First identify its radius and center by comparing the equation with standard form of the equation of the circle:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , with radius ‘ r’ and centre $\left(h,k\right)$ .

Comparing above with ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ gives

  $\left(h,k\right)=\left(0,0\right)\text{ and }r=\sqrt{40}$

Thus, the given circle has centre at $\left(0,0\right)$ and has radius of $2\sqrt{10}$ units.

Thus, the graph of this circle will be as

  

Thus, the radius of circle is $2\sqrt{10}$ units with centre $\left(0,0\right)$ .