Linear Inequalities

Question
CBSEENMA11015476

Let C be the circle with centre at(1,1) and radius 1. If T is the circle centred at (0,y) passing through origin and touching the circle externally, then the radius of T is equal to

  • fraction numerator square root of 3 over denominator square root of 2 end fraction
  • fraction numerator square root of 3 over denominator 2 end fraction
  • 1/2

  • 1/4

Solution

D.

1/4

Let the coordinate of the centre of T be (0, K)
Distance between their centre
straight k plus 1 space equals space square root of 1 plus left parenthesis straight k minus 1 right parenthesis squared end root space left square bracket space because space straight C subscript 1 straight C subscript 2 space equals space straight k plus 1 right square bracket
rightwards double arrow straight k plus 1 space equals space square root of 1 plus straight k squared plus 1 minus 2 straight k end root
rightwards double arrow straight k plus 12 space equals space square root of straight k squared plus 2 minus 2 straight k end root
straight k squared space plus 1 plus 2 straight k space equals space straight k squared space plus 2 minus 2 straight k
rightwards double arrow space straight k space equals space 1 fourth
So, the radius of circle T is k i.e, 1/4

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