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TS EAMCET · Maths · Circle

If the circles \(x^2+y^2-2 x-2(3+\sqrt{7}) y+8+6 \sqrt{7}=0\) and \(x^2+y^2-8 x-6 y+k^2=0, k \in \mathbf{Z}\), have exactly two common tangents, then the number of possible values of \(k\) is

  1. A 8
  2. B 5
  3. C 9
  4. D 11
Verified Solution

Answer & Solution

Correct Answer

(C) 9

Step-by-step Solution

Detailed explanation

The centres of the given circles are \(C_1(1,3+\sqrt{7})\) and \(C_2(4,3)\) and corresponding radii are \(r_1=\sqrt{1^2+(3+\sqrt{7})^2-(8+6 \sqrt{7})}=3\) and \(r_2=\sqrt{4^2+3^2-k^2}=\sqrt{25-k^2}\) Now, \(C_1 C_2=\sqrt{(4-1)^2+(3-3-\sqrt{7})^2}=4\) Clearly,…