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COMEDK2023Evening ShiftMathematicsCircleActual

The centre of the circle passing through (0,0) and (1,0) and touching the circle x^2+y^2=9 is

Options

  1. A( 1 2 , 3 2 )
  2. B( 3 2 , 1 2 )
  3. C( 1 2 ,- 2 )
  4. D( 1 2 , 1 2 )

Correct answer

C. ( 1 2 ,- 2 )

Step-by-step solution

Centre lies on perpendicular bisector of (0,0) and (1,0) , which is x = 1 2 . So h = 1 2 . r^2 = h^2 + k^2 = 1 4 + k^2 Distance between centres C₁ = ( 1 2 , k ) and C₂ = (0,0) : d = 1 4 + k^2 For the circles to touch internally: d = R - r 1 4 +k^2 = 3 - 1 4 +k^2 2 1 4 +k^2 = 3 1 4 +k^2 = 3 2 1 4 + k^2 = 9 4 k^2 = 2 k = 2 Centre = ( 1 2 , - 2 )

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