From a point A(0,3) on the circle (x+2)^2+(y-3)^2=4 , a chord A B...

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