The internal centre of similitude of the two | vedclass

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(C) The given circles are:$x^2+y^2-4x-6y+12=0 \dots(1)$Comparing with $x^2+y^2+2gx+2fy+c=0$,we get center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3

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$\left(1, \frac{5}{2}\right)$

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(C) The given circles are:$x^2+y^2-4x-6y+12=0 \dots(1)$Comparing with $x^2+y^2+2gx+2fy+c=0$,we get center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-12} = \sqrt{13-12} = 1$.$x^2+y^2+4x-2y-4=0 \dots(2)$Comparing with $x^2+y^2+2gx+2fy+c=0$,we get center $C_2 = (-2, 1)$ and radius $r_2 = \sqrt{(-2)^2+1^2-(-4)} = \sqrt{4+1+4} = 3$.The internal centre of similitude divides the line segment joining the centers $C_1$ and $C_2$ internally in the ratio $r_1 : r_2 = 1 : 3$.Using the section formula,the coordinates are:$C = \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)$$C = \left(\frac{1(-2) + 3(2)}{1+3}, \frac{1(1) + 3(3)}{1+3}\right)$$C = \left(\frac{-2+6}{4}, \frac{1+9}{4}\right) = \left(\frac{4}{4}, \frac{10}{4}\right) = \left(1, \frac{5}{2}\right)$.

The internal centre of similitude of the two circles $x^2+y^2-4x-6y+12=0$ and $x^2+y^2+4x-2y-4=0$ is

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