For the given circles x^2 + y^2 - 6x - 2y + 1 | vedclass

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(D) For the first circle $x^2 + y^2 - 6x - 2y + 1 = 0$,the center $C_1 = (3, 1)$ and radius $R_1 = \sqrt{3^2 + 1^2 - 1} = \sqrt{9} = 3$. For the secon

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They touch each other

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(D) For the first circle $x^2 + y^2 - 6x - 2y + 1 = 0$,the center $C_1 = (3, 1)$ and radius $R_1 = \sqrt{3^2 + 1^2 - 1} = \sqrt{9} = 3$. For the second circle $x^2 + y^2 + 2x - 8y + 13 = 0$,the center $C_2 = (-1, 4)$ and radius $R_2 = \sqrt{(-1)^2 + 4^2 - 13} = \sqrt{1 + 16 - 13} = \sqrt{4} = 2$. The distance between the centers is $C_1C_2 = \sqrt{(3 - (-1))^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$. Since $R_1 + R_2 = 3 + 2 = 5$,we have $C_1C_2 = R_1 + R_2$. Therefore,the circles touch each other externally.

For the given circles $x^2 + y^2 - 6x - 2y + 1 = 0$ and $x^2 + y^2 + 2x - 8y + 13 = 0$,which of the following is true?

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