Let r _ 1 and r _ 2 be the radii of the largest and smallest...

We have,

${x}^{2} + {y}^{2} + 2x + 4y - 4 = 0$

$⇒ {x}^{2} + 2x + 1 + {y}^{2} + 4y + 4 = 9$

$⇒ {\left(x + 1\right)}^{2} + {\left(y + 2\right)}^{2} = {3}^{2}$

Centre$≡ \left(- 1, - 2\right)$

$CP = \sqrt{{\left(- 4 + 1\right)}^{2} + {\left(1 + 2\right)}^{2}} = 3\sqrt{2}$

Centre of smallest circle is $A$

Centre of largest circle is $B$.

${r}_{2} = | CP - CA | = 3\sqrt{2} - 3$

${r}_{1} = CP + CB = 3\sqrt{2} + 3$

Hence,

$\dfrac{{r}_{1}}{{r}_{2}} = \dfrac{3\sqrt{2} + 3}{3\sqrt{2} - 3} = \dfrac{( 3\sqrt{2} + 3{)}^{2}}{9}$

$⇒ \dfrac{{r}_{1}}{{r}_{2}} = ( \sqrt{2} + 1{)}^{2} = 3 + 2\sqrt{2}$

Hence,

$a = 3, b = 2$

$a + b = 5$