Then: the slope of incident light is -k,

Therefore, the incident ray equation is: y-3=-k(x+2).

Namely: y=-kx+3-2k.

Because the intersection of incident light and reflected light is located on the X axis,

Combine reflected light with incident light.

①y=kx+b

②y=-kx+3-2k

Change ① to get: x = (y-b)/k.

Substituting ②, we get: y=-k[(y-b)/k]+3-2k.

Namely: 2y=b-2k+3.

Should be: y=0,

So: ③b=2k-3

Substituting the reflected light equation into a circle, we get: (x-3) 2+(kx+b-2) 2 = 1.

(k^2+ 1)x^2+(2bk-4k-6)x+b^2-4b+ 12=0

Because the reflected light is tangent to the circle, the above equation has multiple roots, namely:

△=(2bk-4k-6)^2-4(k^2+ 1)(b^2-4b+ 12)=0

After sorting:16k2+12 (b-2) k+(B2-4b-6) = 0.

Substituting in ③ gives:16k2+12 (2k-5) k+(4k2-20k+15) = 0.

After finishing: 44k 2-80k+ 15 = 0.

Solving this equation, we get: k 1=(20+√235)/22, k2=(20-√235)/22.

Substitute: b1= (√ 235-13)/1,B2 =-(13+√ 235)//kloc-0.

Substituting the obtained b and k into the set, the equation of reflected light is:

(20+√235)x-22y+2√235-26=0

(20-√235)x-22y-2√235-26=0