Chapter 8.4, Problem 40E

Solution

Graph the given equation.

The graph is

  

Given information:

Given equation is:

  $30x^2+30y^2=480$

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:

In order to graph the given equation of circle

  $30x^2+30y^2=480$ ,

First simplify it as

  $\begin{array}{c}30x^2+30y^2=480\\ 30\left(x^2+y^2\right)=480\\ x^2+y^2=\frac{480}{30}\\ x^2+y^2=16\\ x^2+y^2=4^2\end{array}$

Now, 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 center $\left(h,k\right)$ .

Rewrite the given equation as:

  $\begin{array}{c}{x}^2+y^2=4^2\\ {\left(x-0\right)}^2+{\left(y-0\right)}^2=4^2\end{array}$

Thus, $\left(h,k\right)=\left(0,0\right)\text{ and }r=4$

Thus, the given equation is of circle with center at origin and radius of units 4.

Thus, the graph of given equation will be as