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

we get

(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

K =

+(8/28)^(1/2)

-(8/28)^(1/2)