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Diameter ends are (1, 2) and (5, 6). What is the circle?
Ax²+y²−6x−8y+17=0
Bx²+y²−6x−8y+7=0
Cx²+y²+6x+8y+17=0
Dx²+y²−2x−4y+17=0
Answer & Solution
Correct answer: A. x²+y²−6x−8y+17=0
Diameter form gives (x − 1)(x − 5) + (y − 2)(y − 6) = 0, which expands to x² + y² − 6x − 8y + 17 = 0.
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