Let be four distinct points on a circle where , . Find .
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Let the equation be of the circle be x 2 + y 2 + 2 g x + 2 f y + c = 0
Since ( m k , m k 1 ) lies on it , it satisfies the equation.
Rearranging the equation we get, m k 4 + 2 g m k 3 + c m k 2 + 2 f m k + 1 = 0
Clearly it's roots are m 1 , m 2 , m 2 , m 4
So ∏ k = 1 4 m k = m 1 m 2 m 3 m 4 = 1 [By vieta's]