Locus problem need help 
My try 
Let $(x,y)$ place on the demanded graph 
$\sqrt {x^{2}+y^{2}}-1=\sqrt {x^{2}+(y+3)^{2}}$
$-2\sqrt {x^{2}+y^{2}}= 6y+8$
$-\sqrt {x^{2}+y^{2}}=3y+4$
$x^{2}+y^{2}= 16+9y^2+24y$
What  should I do now? 

Let $P(h,k)$ be the point which is equidistant (call this distance $d$) from the circle and the line. Then the distance of $P$ from the center of the circle (which in this case is $(0,0)$) should be $d+1$ and since $d$ is also the distance (perpendicular distance) from the horizontal line $y=-3$, so $d=k+3$. 

Thus we have
\begin{align*}
(k+3)+1 & = \sqrt{h^2+k^2}\\
8k+16&=h^2
\end{align*}
Thus the locus is given by $x^2=8y+16$. This is same as $x^2+y^2=(y+4)^2$. (option (d)).
The final locus will look like

