There are two circles whose equation are `x^2+y^2=9` and `x^2+y^2-8x-6y+n^2=0,n in Zdot` If the two circles have exactly two common tangents, then the number of possible values of `n` is 2 (b) 8 (c) 9 (d) none of these

Options

A. 2
B. 8
C. 9
D. None of these

Answer

3

Explanation

For `x^(2) + y^(2)= 9` the centre = (0,0) and their r=3
For `x^(2)+ y^(2) - 8x -6y +n^(2) =0`,
the centre =(4,3) and the radius = `sqrt(4^(3) + 3^(3) -n^(2))`
`therefore 4^(2) + 3^(2) - n^(2) gt 0` or `n^(2) lt 5^(2) or -5lt nlt 5`
Cicrles should cut to have exactly two common tangents.
So, `r_(1) + r_(2) gtd" "["distance between centre"]`
`therefore 3+ sqrt(25-n^(2)) gt sqrt(4^(2) - 3^(2)) or sqrt(25-n^(2))gt2 or 25-n^(2) gt 4`.
`therefore n^(2) lt 21 or- sqrt(21) lt n lt sqrt(21)`
Therefore, common values of n should satisfy `-sqrt(21) lt n lt sqrt(21)`
But `n in Z`,So, `n=-4, -3,....4`