hi, im new but i'll try to help with my knowledge even though its not very good so correct me if i am wrong.
if the line y=kx+3 is a tangent to the circle, then it has one root.
so solving y=kx+3 simultaneously with y^2 + x^2 = 7
(kx+3)^2 + x^2 = 7
which simplifies to
(x^2)(K^2 +1) + 6kx + 2 = 0
a=k^2 +1 b=6k c=2
so if we take the discriminant.
using b^2 - 4ac
(6k)^2 - 8(K^2 +1)
28K^2 - 8 = 0
K^2 = 8/28