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(B) The given circles are $C_1: x^2+y^2=3$ and $C_2: x^2+y^2-2x+4y+4=0$. For $C_1$,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{3}$. For $C_2$,th

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$-2$

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(B) The given circles are $C_1: x^2+y^2=3$ and $C_2: x^2+y^2-2x+4y+4=0$. For $C_1$,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{3}$. For $C_2$,the center $C_2 = (1, -2)$ and radius $r_2 = \sqrt{1^2+(-2)^2-4} = \sqrt{1+4-4} = 1$. The external centre of similitude divides the line segment joining the centers in the ratio $r_1 : r_2$ externally. Let the external centre be $(\alpha, \beta) = \left( \frac{r_1 x_2 - r_2 x_1}{r_1 - r_2}, \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2} \right)$. Substituting the values: $\alpha = \frac{\sqrt{3}(1) - 1(0)}{\sqrt{3}-1} = \frac{\sqrt{3}}{\sqrt{3}-1}$ and $\beta = \frac{\sqrt{3}(-2) - 1(0)}{\sqrt{3}-1} = \frac{-2\sqrt{3}}{\sqrt{3}-1}$. Therefore,$\frac{\beta}{\alpha} = \frac{-2\sqrt{3}}{\sqrt{3}-1} \div \frac{\sqrt{3}}{\sqrt{3}-1} = -2$.

If $(\alpha, \beta)$ is the external centre of similitude of the circles $x^2+y^2=3$ and $x^2+y^2-2x+4y+4=0$,then $\frac{\beta}{\alpha}=$

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