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$C_1$ and $C_2$ are the external and internal centres of similitude of the circles $x^2+y^2-2 x+4 y+1=0$ and $x^2+y^2+4 x-6 y+12=0$. If the radius of the circle having $C_1 C_2$ as its diameters is $r$, then $\frac{9}{2} r=$
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The correct answer is:
$3 \sqrt{34}$
$C_1$ and $C_2$ are external and internal similitude of the circle
$x^2+y^2-2 x+4 y+1=0$
and $\quad x^2+y^2+4 x-6 y+12=0$
Centre of circle $x^2+y^2-2 x+4 y+1=0$ is $(1,-2), r_1=2$
Centre of circle $x^2+y^2+4 x-6 y+4=0$ $(-2,3), r_2=1$
Coordinate of $C_1=\frac{-5}{3}$
$\left(\frac{2(-2)-1}{2-1}, \frac{6+2}{2-1}\right) \equiv(-5,8)$
Coordinate of $C_2=\left(\frac{2(-2)+1}{2+1}, \frac{6-2}{2+1}\right) \equiv\left(-1, \frac{4}{3}\right)$
$C_1 C_2=\sqrt{(-5+1)^2+\left(8-\frac{4}{3}\right)^2}$
$C_1 C_2=\sqrt{16+\frac{400}{9}} \Rightarrow C_1 C_2=\sqrt{\frac{544}{9}}=\frac{4 \sqrt{34}}{3}$
Here, $C_1 C_2=2 r$
$\therefore \quad \frac{4}{3} \sqrt{34}=2 r$
$\frac{9}{2} r=3 \sqrt{34}$
$x^2+y^2-2 x+4 y+1=0$
and $\quad x^2+y^2+4 x-6 y+12=0$
Centre of circle $x^2+y^2-2 x+4 y+1=0$ is $(1,-2), r_1=2$
Centre of circle $x^2+y^2+4 x-6 y+4=0$ $(-2,3), r_2=1$
Coordinate of $C_1=\frac{-5}{3}$
$\left(\frac{2(-2)-1}{2-1}, \frac{6+2}{2-1}\right) \equiv(-5,8)$
Coordinate of $C_2=\left(\frac{2(-2)+1}{2+1}, \frac{6-2}{2+1}\right) \equiv\left(-1, \frac{4}{3}\right)$
$C_1 C_2=\sqrt{(-5+1)^2+\left(8-\frac{4}{3}\right)^2}$
$C_1 C_2=\sqrt{16+\frac{400}{9}} \Rightarrow C_1 C_2=\sqrt{\frac{544}{9}}=\frac{4 \sqrt{34}}{3}$
Here, $C_1 C_2=2 r$
$\therefore \quad \frac{4}{3} \sqrt{34}=2 r$
$\frac{9}{2} r=3 \sqrt{34}$
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