jmc

Solving for $y$ in $5x+8y=10,$ we find $y=\frac{10-5x}{8}.$ Substituting into $x^2+y^2=1,$ we get $$x^2+{\left(\frac{10-5x}{8}\right)}^2=1.$$This simplifies to $89x^2-100x+36=0.$ The discriminant of this quadratic is $100^2-4\cdot 89\cdot 36=-2816.$ Since the discriminant is negative, the quadratic has no real roots. Therefore, the line and circle intersect at $\boxed{0}$ points.