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JEE MATHEMATICS MCQ: CIRCLES

1 Easy
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Application +4 | -1
The inverse of the point (1,2)\left( {1,2} \right) with respect to the circle x2+y2−4x−6y+9=0{x^2} + {y^2} - 4x - 6y + 9 = 0, is
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Correct option: C
(0,1)\left( {0,1} \right)
The correct answer is D ((0,1)) because finding the foot of the perpendicular from the point (1,2) to the tangent line of the circle at that point gives the inverse point.

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