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 
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