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JEE Math Practice Question

If a line, y = mx + c is a tangent to the circle, (x – 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2 + y2 = 1 at the point $$\left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$$, then :

  1. A.c2 + 6c + 7 = 0
  2. B.c2 - 7c + 6 = 0
  3. C.c2 – 6c + 7 = 0
  4. D.c2 + 7c + 6 = 0

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