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