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.
28.