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JEE MATHEMATICS MCQ: CIRCLES

1 Easy
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Application +4 | -1
The tangent at (1,7)\left( {1,7} \right) to the curve x2=y−6{x^2} = y - 6 touches the circle x2+y2+16x+12y+c=0{x^2} + {y^2} + 16x + 12y + c = 0 at
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Correct option: D
(−6,−7)\left( { - 6, - 7} \right)
Option D ((-6, -7)) is correct because the tangent to the curve at (1,7) touches the circle at (-6,-7) after solving the equations of the tangent and the circle.

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