1. Parabola and endpoints of latus rectum

For the parabola $y^2=4x,$ we compare with the standard form $y^2=4ax$, so $a=1.$

For $y^2=4ax$, the endpoints of the latus rectum are $(a,2a) \quad \text{and} \quad (a,-2a).$ Thus here the endpoints are $(1,2) \quad \text{and} \quad (1,-2).$

2. Equation of normal to $y^2=4ax$ at parameter point

A parametric point on $y^2=4ax$ is $(at^2,2at).$ For this parabola, the normal at parameter $t$ is $y=-tx+2at+at^3.$ Since $a=1$, this becomes $y=-tx+2t+t^3.$

3. Normals at the endpoints of latus rectum

The endpoints $(1,2)$ and $(1,-2)$ correspond to parameters $t=1 \quad \text{and} \quad t=-1,$ because $(t^2,2t)=(1,2), (1,-2).$

At $t=1$:

$y=-x+2+1=-x+3$ so the normal is $x+y-3=0.$

At $t=-1$:

$y=-(-1)x+2(-1)+(-1)^3=x-2-1=x-3$ so the normal is $x-y-3=0.$

4. Condition for a line to be tangent to the circle

The circle is $(x-3)^2+(y+2)^2=r^2,$ so its center is $(3,-2).$

If a line is tangent to the circle, then the perpendicular distance from the center to the line equals the radius $r$.

We are told that both normals are tangents to the circle. So the distance from $(3,-2)$ to either line must be $r$.

5. Distance from center to first normal

Line: $x+y-3=0.$ Distance from $(3,-2)$ is $d_1=\frac{|3+(-2)-3|}{\sqrt{1^2+1^2}}=\frac{|-2|}{\sqrt2}=\sqrt2.$ Thus $r=\sqrt2.$

6. Check with second normal

Line: $x-y-3=0.$ Distance from $(3,-2)$ is $d_2=\frac{|3-(-2)-3|}{\sqrt{1^2+(-1)^2}}=\frac{|2|}{\sqrt2}=\sqrt2.$ Again, $r=\sqrt2.$

So $r^2=2.$

7. Comparison with stored answer

Our derived answer is $\boxed{2}.$ The stored correct answer is also $2$.

Hence, the answer agrees with the stored correct answer.