C_1 and C_2 are the external and internal cen | vedclass

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(D) For the circle $S_1: x^2+y^2-2x+4y+1=0$,the centre $O_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2+(-2)^2-1} = 2$.For the circle $S_2: x^2+y^2+4x-6y+1

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$3\sqrt{34}$

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(D) For the circle $S_1: x^2+y^2-2x+4y+1=0$,the centre $O_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2+(-2)^2-1} = 2$.For the circle $S_2: x^2+y^2+4x-6y+12=0$,the centre $O_2 = (-2, 3)$ and radius $r_2 = \sqrt{(-2)^2+3^2-12} = 1$.The external centre of similitude $C_1$ divides $O_1O_2$ externally in the ratio $r_1:r_2 = 2:1$.$C_1 = \left(\frac{2(-2)-1(1)}{2-1}, \frac{2(3)-1(-2)}{2-1}ight) = (-5, 8)$.The internal centre of similitude $C_2$ divides $O_1O_2$ internally in the ratio $r_1:r_2 = 2:1$.$C_2 = \left(\frac{2(-2)+1(1)}{2+1}, \frac{2(3)+1(-2)}{2+1}ight) = \left(-1, \frac{4}{3}ight)$.The diameter of the circle is the distance $C_1C_2 = \sqrt{(-5 - (-1))^2 + (8 - 4/3)^2} = \sqrt{(-4)^2 + (20/3)^2} = \sqrt{16 + 400/9} = \sqrt{544/9} = \frac{4\sqrt{34}}{3}$.Since $C_1C_2$ is the diameter,$2r = \frac{4\sqrt{34}}{3}$,so $r = \frac{2\sqrt{34}}{3}$.Therefore,$\frac{9}{2}r = \frac{9}{2} 	imes \frac{2\sqrt{34}}{3} = 3\sqrt{34}$.

$C_1$ and $C_2$ are the external and internal centres of similitude of the circles $x^2+y^2-2x+4y+1=0$ and $x^2+y^2+4x-6y+12=0$. If the radius of the circle having $C_1C_2$ as its diameter is $r$,then $\frac{9}{2}r=$

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