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Easy

Question

If a circle passes through the point (a, b) and cuts the circle x to the power of 2 end exponent plus y to the power of 2 end exponent equals p to the power of 2 end exponent orthogonally, then the equation of the locus of its centre is‐

  1. x to the power of 2 end exponent plus y to the power of 2 end exponent minus 3 a x minus 4 b y plus left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent minus p to the power of 2 end exponent right parenthesis equals 0    
  2. 2 a x plus 2 b y minus left parenthesis a to the power of 2 end exponent minus b to the power of 2 end exponent plus p to the power of 2 end exponent right parenthesis equals 0    
  3. x to the power of 2 end exponent plus y to the power of 2 end exponent minus 2 a x minus 3 b y plus left parenthesis a to the power of 2 end exponent minus b to the power of 2 end exponent minus p to the power of 2 end exponent right parenthesis equals 0    
  4. 2 a x plus 2 b y minus left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent plus p to the power of 2 end exponent right parenthesis equals 0    

The correct answer is: 2 a x plus 2 b y minus left parenthesis a to the power of 2 end exponent plus b to the power of 2 end exponent plus p to the power of 2 end exponent right parenthesis equals 0


    To find locus for a given condition

    Let the circle be x2+y2+2gx+2fy+c=0.

    ince it cuts x2+y2k2=0 orthogonally therefore
    2g(0)+2f(0)=ck2=0c=k2
    Again the circle C2 passes through (a,b)
    a2+b2+2ga+2fb+k2=0

    c=k2
     Locus of centre (g,f) is
    a2+b2+2a(x)+2b(y)+k2=0
    or 2ax+2by(a2+b2+k2)=0.

    Hence locus of the given condition is optionD

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