TS EAMCET 4 May 2018 Shift 1 Solved Paper

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.