From a point $\mathrm{A}(0,3)$ 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

• Correct answer
  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

  $(x+1)^2+(y-3)^2=1$

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 $\mathrm{AQ}=2 \mathrm{AB}$. Then, $\mathrm{B}=\left(\frac{0+h}{2}, \frac{k+3}{2}\right) \rightarrow\left(\frac{h}{2}, \frac{k+3}{2}\right)$ Point B also satisfy the equation of circle. $(x+2)^2+(y-3)^2=4$ $\left(\frac{h}{2}+2\right)^2+\left(\frac{k+3}{2}-3\right)^2=4$ $\frac{(h+4)^2}{4}+\frac{(k-3)^2}{4}=4$ $(h+4)^2+(k-3)^2=16$ Replace $(h, k)$ by $(x, y)$, then, the required equation is $(x+4)^2+(y-3)^2=16$

Hence, the correct answer is option: 'A'.