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What is the equation of the circle through (5, 7), (6, 6) and (2, −2)?
Ax² + y² + 4x + 6y − 12 = 0
Bx² + y² − 4x − 6y + 12 = 0
Cx² + y² − 6x − 4y − 12 = 0
Dx² + y² − 4x − 6y − 12 = 0
Answer & Solution
Correct answer: D. x² + y² − 4x − 6y − 12 = 0
Substituting the three points gives g = −2, f = −3, c = −12, and each point satisfies x² + y² − 4x − 6y − 12 = 0.
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