Locus of a point whose circle intersects given two circles.

From $x_E=-x_G,y_E=-y_G,JE=JG$, it follows that:

$$(x_J-x_G)^2+(y_J-y_G)^2=(x_J+x_G)^2+(y_J+y_G)^2$$

This equation can be easily simplified to:

$$x_Gx_J+y_Gy_J=0\tag{1}$$

From $x_H=5+(5-x_F)=10-x_F, y_H=-y_F, JF=JH$, it follows that:

$$(x_J-x_F)^2+(y_J-y_F)^2=(x_J-(10-x_F))^2+(y_J+y_F)^2$$

This equation can be simplified to:

$$25-5x_F+x_Fx_J+y_Fy_J=5x_J\tag{2}$$

It is also obvious that:

$$x_G^2+y_G^2=4\tag{3}$$

$$(x_F-5)^2+y_F^2=1$$

The last equation can be written as:

$$x_F^2+y_F^2=10x_F-24 \tag{4}$$

Finally, we have that JF=JG:

$$(x_J-x_F)^2+(y_J-y_F)^2=(x_J-x_G)^2+(y_J-y_G)^2\tag{5}$$

This is it. This is our system that has 6 unknowns and 5 equations. Let's see what we can extract from that...

After a few elementary simplifications, equation (5) becomes:

$$x_F^2+y_F^2+2x_Gx_J-2y_Fy_J+2y_Gy_J-2x_Fx_J=x_G^2+y_G^2\tag{6}$$

If you put (3) and (4) into (6), you get:

$$10x_F-24+2x_Gx_J-2y_Fy_J+2y_Gy_J-2x_Fx_J=4\tag{7}$$

$$10x_F-24+2(x_Gx_J+y_Gy_J)-2y_Fy_J-2x_Fx_J=4\tag{8}$$

Combine (1) and (8) and you get:

$$10x_F-2y_Fy_J-2x_Fx_J=28\tag{9}$$

Multiply equation (2) by 2 and you get:

$$50-10x_F+2x_Fx_J+2y_Fy_J=10x_J\tag{10}$$

Here is the magic: add equations (9) and (10) and you get:

$$50=28+10x_J\implies10x_J=22$$

$$\boxed{x_J=\frac{11}5}\tag{11}$$

So we proved what the value of $x_J$ has to be. Now we can pick any 4 equations from the system (1-5) using the calculated value of $x_J$ and pick one coordinate freely ($y_J$) to calculate the remaining four values (coordinates of points $G$ and $F$).

So your locus of points is a straight line defined by equation (11).

Just for fun, let's see how far we can get without invoking coordinates. We know that the distance between the points $O_1$ and $O_2$ is $5.$ The radii of circles $O_1$ and $O_2$ are respectively $2$ and $1.$ A circle about a point $O$ passes through points $A,B$ on circle $O_1$ and $C,D$ on circle $O_2$ such that $AB=4$ and $CD=2.$ Since the diameter of circle $O_1$ is $4,$ we know that $AB$ is a diameter of circle $O_1.$ For similar reasons, we know $CD$ is a diameter of circle $O_2.$ This leads to the figure below:

Since $O$ is on the perpendicular bisector of any chord of the circle about $O,$ we have right triangles $OO_1A$ and $OO_2D$ with right angles at $O_1$ and $O_2.$ Therefore $$ \begin{align} O_1O^2 &= OA^2 - O_1A^2, \\ O_2O^2 &= OD^2 - O_2D^2, \end{align} $$ and since $OA = OD,$ $O_1A = 2,$ and $O_2D = 1,$ $$ O_2O^2 - O_1O^2 = O_2D^2 - O_1A^2 = 3. \tag1 $$

Now drop a perpendicular from $O$ to $O_1O_2$ meeting $O_1O_2$ at $M.$ We have $$ \begin{align} O_1O^2 &= OM^2 + O_1M^2, \\ O_2O^2 &= OM^2 + O_2M^2, \end{align} $$ from which it follows that$$ O_2O^2 - O_1O^2 = O_2M^2 - O_1M^2. \tag2 $$ From Equations $(1)$ and $(2)$ we then have $$ \begin{align} 3 &= O_2M^2 - O_1M^2 \\ &= (O_2M + O_1M)(O_2M - O_1M) \\ &= 5(O_2M - O_1M). \end{align} $$

Therefore $O_2M - O_1M = \dfrac35.$ Using the fact that $O_2M + O_1M = 5$ again, a little algebra tells us that $O_1M = \dfrac{11}{5}$ and $O_2M = \dfrac{14}{5}.$

But again $O_1M = \dfrac{11}{5}$ and $O_2M = \dfrac{14}{5}$ whenever we place $O$ (aka $P$) to satisfy the given conditions. That is, the locus of $P$ is the set of all points such that the foot of the perpendicular from $P$ to $O_1O_2$ is the same point $M.$

In other words, the locus of $P$ is the line perpendicular to $O_1O_2$ intersecting $O_1O_2$ at $M$ such that $O_1M = \dfrac{11}{5}$ and $O_2M = \dfrac{14}{5}.$

Now if you want to set up a coordinate system with $O_1$ at $(0,0)$ and $O_2$ at $(5,0),$ you'll find that $P$ lies on the line $x = \dfrac{11}{5},$ if that matters. (It wasn't clear from the problem statement whether the answer has to given in $x$ and $y$ coordinates.)