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27) prove that a point can be found which is at the same distant from each of four points: $(a \cdot m_1, \dfrac {a}{m_1}),(a \cdot m_2, \dfrac {a}{m_2}),(a \cdot m_3, \dfrac {a}{m_3})$ and $(\dfrac {a}{m_1 \cdot m_2 \cdot m_3}, a \cdot m_1 \cdot m_2 \cdot m_3)$.
For the first 3 points, they will lie on the circle:-
$$x^2 + y^2 + 2gx + 2fy + c = 0$$ for some g, f and c.
Let $(a \cdot m, \dfrac {a}{m})$ be a point on this circle. Then, $$(am)^2 + (\dfrac {a}{m})^2 + 2g(am) +2f(\dfrac {a}{m}) +c = 0$$
After simplification, we have
$$m^4 + (\dfrac {2g}{a})m^3 + (\dfrac {c}{a^2})m^2 + (\dfrac {2f}{a})m + 1 = 0 $$
Clearly, $m_1$, $m_2$, and $m_3$ are the three roots of the equation.
Since the equation is of degree $4$ in $m$, we can let the fourth root be M and it has to be a point of the form $(a \cdot M, \dfrac {a}{M})$.
Note that, from the theory of equation, the product of these $4$ roots must equal to $1$, (the $x^0$ term). Therefore,
$$m_1 \cdot m_2 \cdot m_3 \cdot M = 1$$
This means $M = \dfrac {1}{ m_1 \cdot m_2 \cdot m_3}$
The point required is $(-g, -f)$, the center of the circle that passes through at least the first three points.