Question
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
BITSAT
Mathematics
Circle
2024
BITSAT 2024 (Memory Based Paper 1)
BITSAT Mathematics Question (2024) — Solution
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
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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