Answer:
[tex](x+3)^{2}+(y+5)^{2}=36[/tex]
Step-by-step explanation:
we know that
The equation of a circle in standard form is equal to
[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
In this problem we have
[tex](h,k)=(-3,-5)[/tex]
[tex]r= 6\ units[/tex]
substitute
[tex](x+3)^{2}+(y+5)^{2}=6^{2}[/tex]
[tex](x+3)^{2}+(y+5)^{2}=36[/tex]