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JEE Main2014MathematicsCircleActual

If the point 1 , 4 lies inside the circle x 2 + y 2 - 6 x + 10 y + p = 0 and the circle does not touch or intersect the coordinate axes, then the set of all possible values of p is the interval

Options

  1. A25 ,   39
  2. B25 ,   29
  3. C0 ,   25
  4. D9 ,   25

Correct answer

B. 25 ,   29

Step-by-step solution

x 2 + y 2 - 6 x - 1 0 y + p = 0 ∴ Centre is 3 ,   5 and radius is 34 - p . ∵   1 ,   4 lies inside the circle. So, 1 + 16 - 6 - 40 + p < 0 ⇒ p < 29             . . . 1 Circle neither touches nor cut the coordinate axes. So, radius of circle 34 - p < 5 (for not touching x - axis ) and 34 - p < 3 (for not touching y - axis) ⇒   9 < p and 25 < p   ⇒   p > 25       . . . 2 From equations

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