For circle x^2+y^2=a^2 and line y=mx+c: substitute to get x^2(1+m^2)+2mcx+(c^2-a^2)=0. Discriminant D = 4m^2c^2 - 4(1+m^2)(c^2-a^2) = 4a^2(1+m^2) - 4c^2. D > 0: secant (two points), D = 0: tangent (one point), D < 0: no intersection. The tangency condition simplifies to c^2 = a^2(1+m^2), equivalently |c|/sqrt(1+m^2) = a.