2.Now, we will check that what type of equation is the r=8

This equation is the polar form equation in which we consider \(\displaystyle{r}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}\)

So, \(\displaystyle{r}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}={8}\)

\(\displaystyle\Rightarrow={x}^{{2}}+{y}^{{2}}={8}^{{2}}={64}\)

Thus, we can say that after changing polar co-ordinates to Cartesian co-ordinates, equation r=8 is also the equation of the circle with radius a=8 and center at (0,0).

So, the graphs of the equation

r=8 and \(\displaystyle{x}^{{2}}+{y}^{{2}}={64}\) are same which is given above.

Now, we have the parametric form equations given as:

\(\displaystyle{x}={8}{\sin{{\left({3}{t}\right)}}},{y}={8}{\cos{{\left({3}{t}\right)}}},{0}\le{t}\le{2}\pi\)

\(\displaystyle{x}^{{2}}={64}{{\sin}^{{2}}{\left({3}{t}\right)}}\)

and \(\displaystyle{y}^{{2}}={64}{{\cos}^{{2}}{\left({3}{t}\right)}}\)

\(\displaystyle\Rightarrow{x}^{{2}}+{y}^{{2}}={64}{{\sin}^{{2}}{\left({3}{t}\right)}}+{64}{{\cos}^{{2}}{\left({3}{t}\right)}}\)

\(\displaystyle={64}{\left({{\sin}^{{2}}{\left({3}{t}\right)}}+{{\cos}^{{2}}{\left({3}{t}\right)}}\right)}={64}\)

as by trigonometry identity \(\displaystyle{{\sin}^{{2}}{\left({3}{t}{0}+{{\cos}^{{2}}{\left({3}{t}\right)}}={1}\right.}}\)

thus, we have shown that these parametric equations from circle with radius 8 and center at (0,0).

So, these parametric equations

\(\displaystyle{x}={8}{\sin{{\left({3}{t}\right)}}},{y}={8}{\cos{{\left({t}\right)}}},{0}\le{t}\le{2}\pi\) also have the same graph given above.

4. So, the given statement is true as the equation

r=8 ,

\(\displaystyle{x}^{{2}}+{y}^{{2}}={64}\) and

\(\displaystyle{x}={8}{\sin{{\left({3}{t}\right)}}},{y}={8}{\cos{{\left({t}\right)}}},{0}\le{t}\le{2}\pi\) denote same equation of circle with radius 8 and center at

origin.Thus, all these equations have the same graph given above.

Answer: it`s TRUE