If C_1 and C_2 are the centres of similitude | vedclass

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(C) For the circle $x^2+y^2+6x+8y+24=0$,the center $O_1 = (-3, -4)$ and radius $r_1 = \sqrt{(-3)^2 + (-4)^2 - 24} = \sqrt{9+16-24} = 1$. For the circl

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$\frac{16}{3}$

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(C) For the circle $x^2+y^2+6x+8y+24=0$,the center $O_1 = (-3, -4)$ and radius $r_1 = \sqrt{(-3)^2 + (-4)^2 - 24} = \sqrt{9+16-24} = 1$. For the circle $x^2+y^2-6x-8y+9=0$,the center $O_2 = (3, 4)$ and radius $r_2 = \sqrt{(3)^2 + (4)^2 - 9} = \sqrt{9+16-9} = 4$. The distance between the centers $O_1$ and $O_2$ is $d = \sqrt{(3 - (-3))^2 + (4 - (-4))^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = 10$. The distance between the centers of similitude $C_1$ and $C_2$ is given by the formula $d 	imes \left| \frac{r_1+r_2}{r_1-r_2} - \frac{r_1-r_2}{r_1+r_2} ight|$ or simply by calculating the coordinates. Alternatively,the distance between the internal and external centers of similitude is $\frac{2 d r_1 r_2}{|r_1^2 - r_2^2|}$. Substituting the values: $d = 10, r_1 = 1, r_2 = 4$. Distance $= \frac{2 	imes 10 	imes 1 	imes 4}{|1^2 - 4^2|} = \frac{80}{|1 - 16|} = \frac{80}{15} = \frac{16}{3}$.

If $C_1$ and $C_2$ are the centres of similitude with respect to the circles $x^2+y^2+6x+8y+24=0$ and $x^2+y^2-6x-8y+9=0$,then $C_1C_2=$

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