If the circles C_1: x^2+y^2+2x+4y-20=0 and C_ | vedclass

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(B) Given circles are $C_1: x^2+y^2+2x+4y-20=0$ and $C_2: x^2+y^2+6x-8y+9=0$. For $C_1$,centre $O_1 = (-1, -2)$ and radius $r_1 = \sqrt{(-1)^2 + (-2)^

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$\sqrt{39}$

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(B) Given circles are $C_1: x^2+y^2+2x+4y-20=0$ and $C_2: x^2+y^2+6x-8y+9=0$. For $C_1$,centre $O_1 = (-1, -2)$ and radius $r_1 = \sqrt{(-1)^2 + (-2)^2 - (-20)} = \sqrt{1+4+20} = 5$. For $C_2$,centre $O_2 = (-3, 4)$ and radius $r_2 = \sqrt{(-3)^2 + 4^2 - 9} = \sqrt{9+16-9} = 4$. The distance between centres $d = O_1O_2 = \sqrt{(-3 - (-1))^2 + (4 - (-2))^2} = \sqrt{(-2)^2 + 6^2} = \sqrt{4+36} = \sqrt{40}$. Since $r_1 - r_2 The centre of similitude (external) is the point from which the length of the tangent $l$ to both circles is equal. The length of the tangent $l$ from the external centre of similitude to circle $C_2$ is given by $l = \frac{r_2 \cdot d}{r_1 - r_2}$ is incorrect; the standard formula for the length of the tangent from the external centre of similitude to $C_2$ is $l = \frac{r_2 \cdot d}{r_1 - r_2}$ is for the distance,but the length of the tangent $l$ from the external centre of similitude to $C_2$ is $l = \sqrt{d_{ext}^2 - r_2^2}$. Actually,the length of the tangent $l$ from the external centre of similitude to $C_2$ is $l = \frac{r_2 \sqrt{d^2 - (r_1-r_2)^2}}{r_1-r_2} = \frac{4 \sqrt{40 - 1}}{1} = 4\sqrt{39}$. Wait,the question implies the length of the tangent from the centre of similitude to $C_2$ is $l$. Using the property of the external centre of similitude $S$,$\frac{SO_1}{SO_2} = \frac{r_1}{r_2} = \frac{5}{4}$. $SO_1 = \frac{5}{4} SO_2$. Also $SO_1 - SO_2 = d = \sqrt{40}$. $SO_2(\frac{5}{4} - 1) = \sqrt{40} \implies SO_2 = 4\sqrt{40}$. $l^2 = SO_2^2 - r_2^2 = (4\sqrt{40})^2 - 4^2 = 16(40) - 16 = 16(39)$. $l = 4\sqrt{39}$. Then $\frac{l}{n^2} = \frac{4\sqrt{39}}{2^2} = \sqrt{39}$.

If the circles $C_1: x^2+y^2+2x+4y-20=0$ and $C_2: x^2+y^2+6x-8y+9=0$ have $n$ common tangents and the length of the tangent drawn from the centre of similitude to the circle $C_2$ is $l$,then $\frac{l}{n^2} =$

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