From a point $A\left(0,3\right)$ on the circle ${\left(x+2\right)}^2+{\left(y-3\right)}^2=4$, a chord $AB$ is drawn and it is extended to a point $Q$ such that $AQ=2 AB$. Then the locus of $Q$ is

Given,

From a point $A\left(0,3\right)$ on the circle ${\left(x+2\right)}^2+{\left(y-3\right)}^2=4$, a chord $AB$ is drawn and it is extended to a point $Q\left(h,k\right)$ such that $AQ=2 AB$.

Which means $B\left(x,y\right)$ is the midpoint of $A\left(0,3\right) \& Q\left(h,k\right)$

Now using the midpoint formula we get,

$x=\frac{h}{2} \& y=\frac{k+3}{2}$

Now putting the value of $x \& y$ in circle equation we get,

${\left(\frac{h}{2}+2\right)}^2+{\left(\frac{k+3}{2}-3\right)}^2=4$

Then the locus of $Q$ is ${\left(x+4\right)}^2+{\left(y-3\right)}^2=16$