Let P Q be the common chord of the circles S 1 : x 2 + y 2 + 2 x + 3 y + 1 = 0 and S 2 : x 2 + y 2 + 4 x + 3 y + 2 = 0 , then the perimeter (in units) of the triangle C 1 P Q is equal to where , C 1 = - 1 , - 3 2

Question

Let P Q be the common chord of the circles S 1 : x 2 + y 2 + 2 x + 3 y + 1 = 0 and S 2 : x 2 + y 2 + 4 x + 3 y + 2 = 0 , then the perimeter (in units) of the triangle C 1 P Q is equal to where , C 1 = - 1 , - 3 2, 9 2, 2 2 + 3, 3 2 + 2, 3 2 + 2 2

MARKS App · Accepted Answer

Equation of the common chord is 2 x + 1 = 0 C 1 M = - 2 + 1 2 = 1 2 P M = 9 4 - 1 4 = 2 Length of the common chord = 2 2 Hence, the perimeter of Δ C 1 P Q = 3 2 + 3 2 + 2 2 = 3 + 2 2 units

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