The locus of all points P whose farthest and shortest distances to the circle ${\left(x-a\right)}^2+{\left(y-b\right)}^2={\left(a+b\right)}^2\text{ are 2a,2b}\left(\text{a>b>0}\right)$ is

From P Farthest and shortest distances to a circle with center at C and radius r are CP + r and $|r-CP|$
Given CP + r = 2a
$|r-CP|=2{\mathrm{b}}_{ ,}\mathrm{r}=\mathrm{a}+\mathrm{b},\mathrm{C}=(\mathrm{a},\mathrm{b})$
$\begin{array}{l}(CP+r{)}^2-(r-CP{)}^2=4r\left(CP\right)\\ \Rightarrow CP=a-b\Rightarrow (x-a{)}^2+(y-b{)}^2=(a-b{)}^2\text{ for }P=(x,y)\end{array}$