The circles x^2 + y^2 - 2x - 4y = 0 and x^2 + | vedclass

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(A) For the first circle $x^2 + y^2 - 2x - 4y = 0$,the center $C_1 = (1, 2)$ and radius $R_1 = \sqrt{1^2 + 2^2 - 0} = \sqrt{5}$.For the second circle

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Touch each other internally

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(A) For the first circle $x^2 + y^2 - 2x - 4y = 0$,the center $C_1 = (1, 2)$ and radius $R_1 = \sqrt{1^2 + 2^2 - 0} = \sqrt{5}$.For the second circle $x^2 + y^2 - 8y - 4 = 0$,the center $C_2 = (0, 4)$ and radius $R_2 = \sqrt{0^2 + 4^2 - (-4)} = \sqrt{20} = 2\sqrt{5}$.The distance between the centers $C_1$ and $C_2$ is $C_1C_2 = \sqrt{(1-0)^2 + (2-4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}$.We observe that $|R_2 - R_1| = |2\sqrt{5} - \sqrt{5}| = \sqrt{5}$.Since $C_1C_2 = |R_2 - R_1|$,the circles touch each other internally.

The circles $x^2 + y^2 - 2x - 4y = 0$ and $x^2 + y^2 - 8y - 4 = 0$:

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