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BITSAT2024MathematicsCircleActual

From a point A (0,3) on the circle (x+2)^2+(y-3)^2=4 , a chord A B is drawn and it is extended to a point Q such that A Q=2 A B . Then the locus of Q is

Options

  1. A(x+4)^2+(y-3)^2=16
  2. B(x+1)^2+(y-3)^2=32
  3. C(x+1)^2+(y-3)^2=4
  4. D(x+1)^2+(y-3)^2=1

Correct answer

A. (x+4)^2+(y-3)^2=16

Step-by-step solution

Given equation of circle (x+2)^2+(y-3)^2=4 Let the coordinates of Q is (h, k) . Coordinate of B which is midpoint of AQ because AQ =2 AB . Then, B = ( 0+h 2 , k+3 2 ) ( h 2 , k+3 2 ) Point B also satisfy the equation of circle. (x+2)^2+(y-3)^2=4 ( h 2 +2 )^2+ ( k+3 2 -3 )^2=4 (h+4)^2 4 + (k-3)^2 4 =4(h+4)^2+(k-3)^2=16 Replace (h, k) by (x, y) , then, the required equation is (x+4)^2+(y-3)^2=16

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