jmc

Let the center of the circle be $(a,b),$ and let its radius be $r.$ Either the two circles are externally or internally tangent to the two original circles.

If the circle is externally tangent to both circles, then the distance between the centers is equal to the sum of the radii, giving us $$\begin{array}{rl}(a-4{)}^2+b^2& =(r+1{)}^2,\\ (a+4{)}^2+b^2& =(r+1{)}^2.\end{array}$$Subtracting, we get $16a=0,$ so $a=0.$ Hence, $$16+b^2=(r+1{)}^2.$$Since the circle passes through $(0,5),$ $$(b-5{)}^2=r^2.$$Subtracting the equations $16+b^2=(r+1{)}^2$ and $(b-5{)}^2=r^2,$ we get $$10b-9=2r+1.$$Then $r=5b-5.$ Substituting into $(b-5{)}^2=r^2,$ we get $$(b-5{)}^2=(5b-5{)}^2.$$This simplifies to $24b^2-40b=0,$ so $b=0$ or $b=\frac{40}{24}=\frac{5}{3}.$ If $b=0,$ then $r=-5,$ which is not possible. If $b=\frac{5}{3},$ then $r=\frac{10}{3},$ giving us one externally tangent circle.



If the circle is internally tangent to both circles, then the distance between the centers is equal to the difference of the radii, giving us $$\begin{array}{rl}(a-4{)}^2+b^2& =(r-1{)}^2,\\ (a+4{)}^2+b^2& =(r-1{)}^2.\end{array}$$Subtracting, we get $16a=0,$ so $a=0.$ Hence, $$16+b^2=(r-1{)}^2.$$Since the circle passes through $(0,5),$ $$(b-5{)}^2=r^2.$$Subtracting the equations $16+b^2=(r-1{)}^2$ and $(b-5{)}^2=r^2,$ we get $$10b-9=-2r+1.$$Then $r=5-5b.$ Substituting into $(b-5{)}^2=r^2,$ we get $$(b-5{)}^2=(5-5b{)}^2.$$This simplifies to $24b^2-40b=0,$ so $b=0$ or $b=\frac{5}{3}.$ If $b=0,$ then $r=5,$ giving us one internally tangent circle. If $b=\frac{5}{3},$ then $r=-\frac{10}{3},$ which is not possible.



Suppose the circle is externally tangent to the circle centered at $(-4,0),$ and internally tangent to the circle centered at $(4,0).$ Then $$\begin{array}{rl}(a+4{)}^2+b^2& =(r+1{)}^2,\\ (a-4{)}^2+b^2& =(r-1{)}^2.\end{array}$$Subtracting these equations, we get $16a=4r,$ so $r=4a.$ Hence, $$(a+4{)}^2+b^2=(4a+1{)}^2.$$Then $b^2=(4a+1{)}^2-(a+4{)}^2=15a^2-15,$ so $a^2=\frac{b^2+15}{15}.$

Since the circle passes through $(0,5),$ $$a^2+(b-5{)}^2=r^2=16a^2.$$Then $(b-5{)}^2=15a^2=b^2+15.$ This gives us $b=1.$ Then $a^2=\frac{16}{15}.$ Since $r=4a,$ $a$ must be positive, so $a=\frac{4}{\sqrt{15}}$ and $r=\frac{16}{\sqrt{15}}.$



By symmetry, there is only one circle that is internally tangent to the circle centered at $(-4,0)$ and externally tangent to the circle centered at $(4,0),$ giving us a total of $\boxed{4}$ circles.