The given circles are, S≡x2+y2−14x+6y+33=0 And, S≡x2+y2−a2=0 For common tangents, the distance between centres of two circles C1,C2 should be more than the sum of radii. Therefore, √49+9>√49+9−33+a √58>√25+a a<√58−5 Now, a∈N Thus, a={1,2} Therefore, two circles are possible.