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JEE MATHEMATICS MCQ: CIRCLES

1 Easy
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Application +4 | -1
The other end of the diameter through the point (−1,1)\left( { - 1,1} \right) on the circle x2+y2−6x+4y−12=0{x^2} + {y^2} - 6x + 4y - 12 = 0 is
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Correct option: C
(7,−5)\left( {7, - 5} \right)
Option A (x2+y2−x−y=0{x^2} + {y^2} - x - y = 0) is correct because it represents the equation of the circle with AB as its diameter, where A and B are the intersection points of the line y=x and the given circle.

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