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TS EAMCET · Maths · Circle

If the angle between the circles \(x^2+y^2-2 x+\mathrm{k} y+1=0\) and \(x^2+y^2-\mathrm{k} x-2 y+1=0\) is \(\operatorname{Cos}^{-1}\left(\frac{1}{4}\right)\) and \(\mathrm{k} < 0\) then the point which lies on the radical axis of the given circles is

  1. A \((1,-3)\)
  2. B \((-1,3)\)
  3. C \((-1,-3)\)
  4. D \((1,3)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \((1,-3)\)

Step-by-step Solution

Detailed explanation

\(r_1 = \sqrt{(-1)^2+(\mathrm{k}/2)^2-1} = |\mathrm{k}/2|\) \(r_2 = \sqrt{(-\mathrm{k}/2)^2+(-1)^2-1} = |\mathrm{k}/2|\) Using \(\operatorname{Cos}\theta = \left|\frac{2g_1g_2+2f_1f_2-c_1-c_2}{2r_1r_2}\right|\)…