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COMEDK · Maths · 12. Circle

\(S \equiv x^2+y^2-2 x-4 y-4=0\) and \(S^{\prime} \equiv x^2+y^2-4 x-2 y-16=0\) are two circles the point \((-2,-1)\) lies

  1. A outside \(S\) and \(S^{\prime}\)
  2. B inside \(S^{\prime}\) only
  3. C inside \(S\) only
  4. D inside \(S\) and \(S^{\prime}\)
Verified Solution

Answer & Solution

Correct Answer

(B) inside \(S^{\prime}\) only

Step-by-step Solution

Detailed explanation

Let the given circles be \(S(x, y) = x^2 + y^2 - 2x - 4y - 4 = 0\) and \(S'(x, y) = x^2 + y^2 - 4x - 2y - 16 = 0\). To determine the position of the point \(P(-2, -1)\) with respect to circle \(S\), substitute the coordinates of \(P\) into the expression for \(S\):…