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) = 4$a^2$(1+$m^2$) - 4$c^2$. 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.