There are two circles x2+y2=9 and x2+y3-8x-6y+n2=0.

Question

There are two circles x2+y2=9 and x2+y2-8x-6y+n2=0. n belongs to z. If the two circles have exactly two common tangents, then the number of possible values of n is

Answer 1

Eqution of 1st circle x²+y2=9, So it'* center is C1u200bu200bu200bu200bu200b->(0,0) and radius r1=3 unit.

Equation of 2nd circle

x²+y2-8x-6y+n2=0

=> (x-4)2+(y-3)2-25+n2=0

So it'* center C2u200bu200bu200bu200bu200b ->(4,3) and radius r2u200bu200bu200bu200bu200b ->(25-n2)^1/2

C1C2=5

C1C2> |r1-r2|

=>5>(25-n2)^1/2

=> |n| < =4

By the given condition n belongs to Z. The possible values of n should satisfy the relation 25-n2>0

=> n = -4,-3,-2,-1,0,+1,+2,+3,+4

So number of possible values of n is 9