Answer
The equation of a circle with center $\left( 3,-5 \right)$ and radius $3$ is${{\left( x-3 \right)}^{2}}+{{\left( y+5 \right)}^{2}}=9$
Work Step by Step
From the figure we see that the circle touches the y-axis at $\left( 0,-5 \right)$,
Since the radius is the distance between $\left( 0,-5 \right)$ and $\left( 3,-5 \right)$,
$\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( -5-\left( -5 \right) \right)}^{2}}}$
$r=3$
Put the value of radius $3$ and center coordinate $\left( 3,-5 \right)$ in the standard equation of the circle,
$\begin{align}
& {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} \\
& {{\left( x-\left( 3 \right) \right)}^{2}}+{{\left( y-\left( -5 \right) \right)}^{2}}={{\left( 3 \right)}^{2}} \\
& {{\left( x-3 \right)}^{2}}+{{\left( y+5 \right)}^{2}}=9
\end{align}$
Therefore, the equation of the circle with center $\left( 3,-5 \right)$ and radius $3$ is ${{\left( x-3 \right)}^{2}}+{{\left( y+5 \right)}^{2}}=9$.
