If $\theta$ is a parameter, then the parametric equations of the circle $x^{2}+y^{2}-6 x+4 y-3=0$ are given by
If $\theta$ is a parameter, then the parametric equations of the circle
$x^{2}+y^{2}-6 x+4 y-3=0$ are given by
$x=-3+4 \sin \theta$ and $y=-2+4 \cos \theta$
$x=3+4 \cos \theta$ and $y=-2+4 \sin \theta$
$x=3+4 \sin \theta$ and $y=2+4 \cos \theta$
$x=3+4 \cos \theta$ and $y=2+4 \sin \theta$
Solution
Given equation of circle is
$\begin{aligned}
& x^{2}+y^{2}-6 x+4 y-3=0 \\
\therefore &\left(x^{2}-6 x+9\right)-9+\left(y^{2}+4 y+4\right)-4-3=0 \\
\therefore &(x-3)^{2}+(y+2)^{2}=16
\end{aligned}$
Comparing with, $(x-h)^{2}+(y-k)^{2}=r^{2}$, we get $h=3, k=-2, r=4$
Parametric form is
$\begin{aligned} x &=h+r \cos \theta & \text { and } & y &=k+r \sin \theta \\ x &=3+4 \cos \theta & \text { and } & y &=-2+4 \sin \theta \end{aligned}$