${\left(\mathrm{x}-2\right)}^2+{\left(\mathrm{y}+3\right)}^2=16$ touching the ellipse $\frac{{\left(\mathrm{x}-2\right)}^2}{{\mathrm{p}}^2}+\frac{{\left(\mathrm{y}+3\right)}^2}{{\mathrm{q}}^2}=1$ from inside. If (2, –6) is one focus of the ellipse then (p, q) =

$\mathrm{S}(2, -6) \mathrm{C}(2, -3)$

$$\begin{gathered} \mathrm{from} \mathrm{the} \mathrm{diagram} \mathrm{clearly} \mathrm{p}=4 \\ and \mathrm{CS}=3 \\ \Rightarrow \mathrm{qe}=3 \\ \Rightarrow {\mathrm{q}}^2{\mathrm{e}}^2=9 \\ \Rightarrow {\mathrm{q}}^2-{\mathrm{p}}^2=9 \\ \Rightarrow {\mathrm{q}}^2=9+16=25 \\ \Rightarrow \mathrm{q}=5 \\ \therefore (\mathrm{p}, \mathrm{q}) =(4, 5) \end{gathered}$$