Solution: The slope perpendicular to the straight line x+y-3=0 is-1, and the slope of the obtained straight line is 1.

Let the linear equation be y=x+b, that is, x-y+b=0.

The circle x 2+y 2-6x-4y+5 = 0 has a center of (3,2) and a radius of two radicals 2.

So there is |3-2+b|/ root number 2=2 root number 2.

That is | 1+b|=4, and the solution is b=3 or b=-5.

So the tangent equation is x-y+3=0 or.

x-y-5=0

Only when the center of the circle is symmetrical about a straight line can the coordinates of the center of the circle be obtained, and the radius is constant, which can be obtained by using the standard equation of the circle.