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