Let $$r_1$$ and $$r_2$$ be the radii of the largest and smallest circles, respectively, which pass through the point $$(-4, 1)$$ and having their centres on the circumference of the circle $$x^2 + y^2 + 2x + 4y - 4 = 0$$. If $$\frac{r_1}{r_2} = a + b\sqrt{2}$$, then $$a + b$$ is equal to:

The given circle is $$x^2 + y^2 + 2x + 4y - 4 = 0$$, which can be rewritten as $$(x+1)^2 + (y+2)^2 = 9$$. This circle has centre $$(-1, -2)$$ and radius $$3$$.

The centres of the required circles lie on this circle, so we parametrise them as $$C = (-1 + 3\cos t,\, -2 + 3\sin t)$$. The radius of each required circle equals $$|CP|$$, where $$P = (-4, 1)$$.

$$|CP|^2 = (-1 + 3\cos t + 4)^2 + (-2 + 3\sin t - 1)^2 = (3 + 3\cos t)^2 + (-3 + 3\sin t)^2$$ $$= 9(1+\cos t)^2 + 9(\sin t - 1)^2 = 9[1 + 2\cos t + \cos^2 t + \sin^2 t - 2\sin t + 1]$$ $$= 9[3 + 2\cos t - 2\sin t].$$

The expression $$2\cos t - 2\sin t$$ has range $$[-2\sqrt{2},\, 2\sqrt{2}]$$, so: $$|CP|^2_{\max} = 9(3 + 2\sqrt{2}), \quad |CP|^2_{\min} = 9(3 - 2\sqrt{2}).$$

Therefore $$r_1 = 3\sqrt{3 + 2\sqrt{2}}$$ and $$r_2 = 3\sqrt{3 - 2\sqrt{2}}$$, giving: $$\frac{r_1}{r_2} = \sqrt{\frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}}} = \sqrt{\frac{(3+2\sqrt{2})^2}{9 - 8}} = 3 + 2\sqrt{2}.$$

Comparing with $$a + b\sqrt{2}$$, we get $$a = 3$$ and $$b = 2$$, so $$a + b = 5$$.