Let r_{1} and r_{2} be the radii of the large | vedclass

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(C) The given circle is $x^{2} + y^{2} + 2x + 4y - 4 = 0$. Its centre $C$ is $(-1, -2)$ and radius $R = \sqrt{(-1)^{2} + (-2)^{2} - (-4)} = \sqrt{1 +

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$5$

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(C) The given circle is $x^{2} + y^{2} + 2x + 4y - 4 = 0$. Its centre $C$ is $(-1, -2)$ and radius $R = \sqrt{(-1)^{2} + (-2)^{2} - (-4)} = \sqrt{1 + 4 + 4} = 3$.Let $P$ be the point $(-4, 1)$. The distance $CP = \sqrt{(-4 - (-1))^{2} + (1 - (-2))^{2}} = \sqrt{(-3)^{2} + 3^{2}} = \sqrt{9 + 9} = 3 \sqrt{2}$.$A$ circle passing through $P$ with centre on the given circle has radius $r = CP = 3 \sqrt{2}$. However,the centre $O$ of such a circle lies on the circle $C$. Thus,the radius $r$ of the circle is the distance between its centre $O$ (which is on the circle $C$) and the point $P$.The distance $r$ varies as $O$ moves on the circle $C$. The minimum distance is $r_{2} = CP - R = 3 \sqrt{2} - 3$ and the maximum distance is $r_{1} = CP + R = 3 \sqrt{2} + 3$.Then $\frac{r_{1}}{r_{2}} = \frac{3 \sqrt{2} + 3}{3 \sqrt{2} - 3} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$.Rationalizing the denominator: $\frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2 + 1 + 2 \sqrt{2}}{2 - 1} = 3 + 2 \sqrt{2}$.Comparing with $a + b \sqrt{2}$,we get $a = 3$ and $b = 2$.Therefore,$a + b = 3 + 2 = 5$.

Let $r_{1}$ and $r_{2}$ be the radii of the largest and smallest circles,respectively,which pass through the point $(-4, 1)$ and have 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:

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