I know that the line and the circle will intersect if and only if there is at least one solution to the system of equations $5x + 8y = 10$ and $x^2 + y^2 = 1$. To find out how many solutions there are, I can try to eliminate one of the variables and solve for the other. For example, I can solve for $x$ in terms of $y$ from the first equation, and then substitute it into the second equation. This gives me $(\frac{10 - 8y}{5})^2 + y^2 = 1$. I can simplify this by multiplying both sides by 25, which gives me $(10 - 8y)^2 + 25y^2 = 25$. Expanding the square, I get $100 - 160y + 64y^2 + 25y^2 = 25$. Combining like terms, I get $89y^2 - 160y + 75 = 0$. This is a quadratic equation in $y$, and I can use the quadratic formula to find its roots. The roots are $y = \frac{160 \pm \sqrt{160^2 - 4(89)(75)}}{2(89)}$. To simplify this expression, I notice that $160^2 - 4(89)(75) = 25600 - 26700 = -1100$. This means that the square root is imaginary, and there are no real values of $y$ that satisfy the equation. Therefore, there are no solutions to the system of equations, and the line and the circle do not intersect at all. 
How many times do the graphs of the line $5x + 8y = 10$ and the circle $x^2 + y^2 = 1$ intersect?