The circles x^2 + y^2 + 2x - 2y + 1 = 0 and x | vedclass

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

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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 both have radii $r_1 = 1$ and $r_2 = 1$.The distance between the centers is $d = C_1C_2 = \sqrt{(1 - (-1))^2 + (1 - 1)^2} = \sqrt{2^2 + 0^2} = 2$.Since the sum of the radii $r_1 + r_2 = 1 + 1 = 2$,and $d = r_1 + r_2$,the circles touch each other externally.The equation of the common tangent is obtained by subtracting the two circle equations:$(x^2 + y^2 + 2x - 2y + 1) - (x^2 + y^2 - 2x - 2y + 1) = 0$$4x = 0 \Rightarrow x = 0$.Substituting $x = 0$ into the first circle equation:$0^2 + y^2 + 2(0) - 2y + 1 = 0$$y^2 - 2y + 1 = 0$ $\Rightarrow (y - 1)^2 = 0$ $\Rightarrow y = 1$.Thus,the point of contact is $(0, 1)$.

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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