AP EAMCET · Maths · Circle
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
- A 8
- B 7
- C 9
- D 4
Answer & Solution
Correct Answer
(C) 9
Step-by-step Solution
Detailed explanation
Círculo 1: \(x^2+y^2=9 \implies C_1=(0,0), R_1=3\) Círculo 2: \(x^2+y^2-8 x-6 y+n^2=0 \implies C_2=(4,3), R_2=\sqrt{4^2+3^2-n^2}=\sqrt{25-n^2}\) Para que \(R_2\) sea real y no degenerado, \(25-n^2 > 0 \implies n^2 Distancia entre centros:…
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