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JEE Main201912 Jan 2019Morning ShiftMathematicsCircleActual

Let C 1 and C 2 be the centres of the circles x 2 + y 2 - 2 x - 2 y - 2 = 0 and x 2 + y 2 - 6 x - 6 y + 14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral P C 1   Q C 2 is :

Options

  1. A6
  2. B4
  3. C8
  4. D9

Correct answer

B. 4

Step-by-step solution

Equation of given circles are x - 1 2 + y - 1 2 = 4 and x - 3 2 + y - 3 2 = 4 . Hence, C 1 1 ,   1 and r 1 = 2 ; C 2 3 ,   3 and r 2 = 2 ⇒ P C 1 = P C 2 = 2 Now, by distance formula, C 1 C 2 = 3 - 1 2 + 3 - 1 2 = 2 2 + 2 2 = 8 ⇒ P C 1 2 + P C 2 2 = C 1 C 2 2 ⇒ ∠ C 1 P C 2 = π 2 (by converse of pythagoras theorem in △ P C 1 C 2 ) Hence, area of quadrilateral P C 1   Q C 2 = 2 × area of △ P C 1 C 2 = 2 × area of △ Q C 1 C 2 = 2 × 1 2 &#21

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