Where do the circles x^2 + y^2 + 2x - 2y + 1 | vedclass

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(A) The centers of the two circles are $C_1(-1, 1)$ and $C_2(1, 1)$,and the radius of both circles is $r_1 = r_2 = 1$.The distance between the centers

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Externally at $(0, 1)$

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(A) The centers of the two circles are $C_1(-1, 1)$ and $C_2(1, 1)$,and the radius of both circles is $r_1 = r_2 = 1$.The distance between the centers is $C_1C_2 = \sqrt{(1 - (-1))^2 + (1 - 1)^2} = \sqrt{2^2 + 0^2} = 2$.Since the sum of the radii is $r_1 + r_2 = 1 + 1 = 2$,and $C_1C_2 = r_1 + r_2$,the circles touch each other externally.To find the point of contact,subtract the two circle equations:$(x^2 + y^2 + 2x - 2y + 1) - (x^2 + y^2 - 2x - 2y + 1) = 0$$4x = 0 \implies x = 0$.Substitute $x = 0$ into either equation:$0^2 + y^2 + 2(0) - 2y + 1 = 0$$y^2 - 2y + 1 = 0 \implies (y - 1)^2 = 0 \implies y = 1$.Thus,the circles touch externally at the point $(0, 1)$.

Where do the circles $x^2 + y^2 + 2x - 2y + 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ touch each other?

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