#### Answer

True

#### Work Step by Step

Given: $x^{2}+y^{2}=4$ and $r=2 $ or $r^{2}=4$
and $x=2sin3t,y=2cos3t$
The Cartesian equation is $x^{2}+y^{2}=r^{2}$
Then
$x^{2}+y^{2}=(2sin3t)^{2}+(2cos3t)^{2}$
$=4sin^{2}3t+4cos^{2}3t$
$=4(sin^{2}3t+cos^{2}3t)$
$=4(1)$
Therefore, $x^{2}+y^{2}=4$
Hence, the given statement is true.