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(D) For the circle $x^2+y^2+2x-6y+1=0$,the centre $C_1 = (-1, 3)$ and radius $r_1 = \sqrt{(-1)^2 + 3^2 - 1} = \sqrt{1+9-1} = 3$.For the circle $x^2+y^

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$5$

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(D) For the circle $x^2+y^2+2x-6y+1=0$,the centre $C_1 = (-1, 3)$ and radius $r_1 = \sqrt{(-1)^2 + 3^2 - 1} = \sqrt{1+9-1} = 3$.For the circle $x^2+y^2-4x+2y+4=0$,the centre $C_2 = (2, -1)$ and radius $r_2 = \sqrt{2^2 + (-1)^2 - 4} = \sqrt{4+1-4} = 1$.The internal centre of similitude $(h, k)$ divides the line segment joining the centres $C_1$ and $C_2$ internally in the ratio $r_1 : r_2 = 3 : 1$.Using the section formula,$(h, k) = \left( \frac{r_1 x_2 + r_2 x_1}{r_1 + r_2}, \frac{r_1 y_2 + r_2 y_1}{r_1 + r_2} \right)$.$(h, k) = \left( \frac{3(2) + 1(-1)}{3+1}, \frac{3(-1) + 1(3)}{3+1} \right) = \left( \frac{6-1}{4}, \frac{-3+3}{4} \right) = \left( \frac{5}{4}, 0 \right)$.Thus,$h = \frac{5}{4}$,which implies $4h = 5$.

If $(h, k)$ is the internal centre of similitude of the circles $x^2+y^2+2x-6y+1=0$ and $x^2+y^2-4x+2y+4=0$,then $4h=$

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