If the circles $x^2+y^2=9$ and $x^2+y^2-8 x-6 y+n^2=0, n \in \mathbb{Z}$ have exactly two common tangents,…

If the circles $x^2+y^2=9$ and $x^2+y^2-8 x-6 y+n^2=0, n \in \mathbb{Z}$ have exactly two common tangents, then the number of values for $n$ is
  1. 8
  2. 7
  3. 9
  4. 4

Solution

No solution. Refer to answer key.

Asked in: AP EAMCET 2017 (25 Apr Shift 1)

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