jmc

By symmetry, the two circles are tangent to each other at the origin $(0,0).$ Therefore, their centers are at the points $(\pm r,0).$ In particular, the circle on the right has the equation $$(x-r{)}^2+y^2=r^2.$$We solve this equation simultaneously with $x^2+5y^2=6.$ Multiplying the first equation by $5$ and subtracting the second equation gives $$[5(x-r{)}^2+5y^2]-[x^2+5y^2]=5r^2-6,$$or $$4x^2-10xr+5r^2=5r^2-6.$$Thus, $$4x^2-10xr+6=0.$$Since the circle on the right and ellipse intersect in two points with the same $x$-coordinate, this quadratic must have exactly one solution for $x.$ Therefore, the discriminant must be zero: $$(10r{)}^2-4\cdot 4\cdot 6=0.$$The positive solution for $r$ is $r=\boxed{\frac{2\sqrt{6}}{5}}.$