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? 
 A: Well, you should start with formulating the "distance to a circle" and "distance to a line", which probably means "distance to the closest point on a ..."
Let's start with the line. The line is horizontal, so only the distance from the y-coordinate to $-3$ matters. This expression becomes the distance between $y$ and $-3$, or $|y+3|$.
Next, the circle. Imagine a line going from $(x, y)$ to the centre of the circle. The distance between $(x, y)$ and the closest point on the circle would be related to the distance between $(x, y)$ and the centre, in fact, the difference between the 2 distances would be exactly $1$. You can formulate this distance as $|\sqrt{x^2+y^2}-1|$.
Hence, now you equate the 2 expressions together.
$|\sqrt{x^2+y^2}-1| = |y+3|$
However, by a simple sketch, you can note that there are no points of the locus in the circle or have $y<-3$. Hence, this reduces to:
$\sqrt{x^2+y^2}-1 = y+3$
By doing a few more manipulations, you can get one of the options above.
A: 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

