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AP EAMCET201920 Apr 2019Morning ShiftMathematicsCircleActual

Two straight rods of lengths (2 a ) and (2 b ) move along the coordinate axes in such a way that their extremities are always concyclic. Then the locus of the centres of such circles is

Options

  1. A(2 (x^2+y^2 )=a^2+b^2 )
  2. B(2 (x^2-y^2 )=a^2+b^2 )
  3. C(x^2+y^2=a^2+b^2 )
  4. D(x^2-y^2=a^2-b^2 )

Correct answer

D. (x^2-y^2=a^2-b^2 )

Step-by-step solution

According to given information, if we draw the figure. Let the equation of circle is ( aligned & x^2+y^2+2 g x+2 f y+c=0 & 2 g^2-c =2 a & and 2 f^2-c =2 b & then g^2-a^2=0 and f^2-b^2=0 & so, g^2-a^2=f^2-b^2 & g^2-f^2=a^2-b^2 aligned ) On taking locus of the centre ((-g,-f) ), we get (x^2-y^2=a^2-b^2 ) Hence, option (4) is correct.

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