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Find the equation of the circle passing through the points of intersection of the circles x² + y² - 4*x - 6*y - 12 = 0 and x² + y² + 6*x + 4*y - 12 = 0, and which cuts the circle x² + y² - 2*x - 4 = 0 orthogonally.

1. x² + y² + 16*x + 14*y - 12 = 0
2. x² + y² - 4*x - 6*y - 12 = 0
3. x² + y² + 6*x + 4*y - 12 = 0
4. x² + y² + 10*x - 2*y - 12 = 0

Correct answer: x² + y² + 16*x + 14*y - 12 = 0

Solution

Writing the family as S1 + lambda*(S1 - S2) = 0 and imposing orthogonality with x²+y²-2x-4=0 gives lambda = -2, yielding x² + y² + 16*x + 14*y - 12 = 0.

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