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

Options

  1. Aoutside S and S^
  2. Binside S^ only
  3. Cinside S only
  4. Dinside S and S^

Correct answer

B. inside S^ only

Step-by-step solution

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 : S(-2, -1) = (-2)^2 + (-1)^2 - 2(-2) - 4(-1) - 4 = 4 + 1 + 4 + 4 - 4 = 9 . Since S(-2, -1) = 9 > 0 , the point P(-2, -1) lies outside the circle S . To determine the position of the point P(-2, -1) with respect to circle S' , substitute the coordinates of P into the expressio

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