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
- 8
- 7
- 9
- 4
Solution
No solution. Refer to answer key.
Asked in: AP EAMCET 2017 (25 Apr Shift 1)
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