If you observe carefully you will find that the given points are not collinear. We will start by considering the first three points. We know that there always exist a circle that pass through any three non collinear points. So the centre of ths circle is equidistant from the three points. Now we will prove that the fourth point also lies on this circle. 

Let us represent the first three points parametrically as (am,a/m). Let the centre of this circle be (α,β). Let the radius of the circle be r. So 

 (am-α)² + (a/m-β)² = r²

So this is biquadratic equation. So it must have four roots. m₁,m₂,m₃ are already roots of this eqn. Let the fourth root be m₄.

If you expand the eqn you will find that the product of the roots is 1. ie m₁m₂m₃m₄ = 1

So we get m₄ = 1/m₁m₂m₃

 ie the fouth point is (am₄,a/m₄) ie (a/m₁m₂m₃,am₁m₂m₃)

Please like the solution if you understand it.

RAJORSHI PAUL

1st year 

Dept of Electrical Engineering

IIT Kharagpur

CONSIDER THAT THESE POINTS LIE ON THE CIRCLE THEN THE POINT WHICH WILL BE EQUIDISTANT FROM ALL THESE POINTS WOULD BE THE CENTER OF THE CIRCLE AND THE DISTANCE WOULD BE THE RADIUS CONSIDER THE CENTER TO BE (a,b) THEN BY USNG DISTANCE FORMULA EQUATE THE DISTANCES BETWEEN (a,b) AND ALL THOSE FOUR POINTS AND GET YOUR ANSWER 

   PLZZZZZZZZ DO APPROVE IF IT HELPED YOU BY CLICKING YES BUTTON

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ APPROVE THE ANSWER IF IT HELPED YOU BY CLICKING THE YES BUTTON...............