Find the shortest distance from the origin to the graph of the circle with equation given below. \[ x^{2}-10 x+y^{2}-12 y+60=0 \]

Step 1 ：The equation of the circle is in the form of a general equation of a circle, which is \(x^2 + y^2 + 2gx + 2fy + c = 0\). The center of the circle is given by the point \((-g, -f)\) and the radius is given by \(\sqrt{g^2 + f^2 - c}\).

Step 2 ：Given the equation of the circle \(x^{2}-10 x+y^{2}-12 y+60=0\), we can identify that \(g = -5\), \(f = -6\), and \(c = 60\).

Step 3 ：Thus, the center of the circle is at the point \((5, 6)\) and the radius of the circle is \(\sqrt{(-5)^2 + (-6)^2 - 60} = 1.0\).

Step 4 ：The shortest distance from the origin to the graph of the circle would be the distance from the origin to the center of the circle minus the radius of the circle. If the result is negative, it means the origin is inside the circle, so the shortest distance is 0.

Step 5 ：The distance from the origin to the center of the circle is \(\sqrt{(5-0)^2 + (6-0)^2} = 7.810249675906654\).

Step 6 ：Subtracting the radius of the circle from the distance to the center gives us the shortest distance from the origin to the circle, which is \(7.810249675906654 - 1.0 = 6.810249675906654\).

Step 7 ：Final Answer: The shortest distance from the origin to the graph of the circle is \(\boxed{6.810249675906654}\).