Question

Prove that a point can be found which is at the same distance from each of the four points
$(a{m}_1,\frac{a}{m_1}),(a{m}_2,\frac{a}{m_2}),(a{m}_3,\frac{a}{m_3})$ and $(a{m}_1{m}_2{m}_3,\frac{a}{m_1{m}_2{m}_3})$

Answer 1

Let $A(a{m}_1,\frac{a}{m_1}),B(a{m}_2,\frac{a}{m_2}),C(a{m}_3,\frac{a}{m_3}),D(a{m}_1{m}_2{m}_2,\frac{a}{m_1{m}_2{m}_3})$

We observe that A,B,C are non-collinear.

So there exists a circle which passes through A,B,C.

Let center of the circle is $(\alpha ,\beta )$

Parametric form of points $A,B,C$ are $(am,\frac{a}{m})$

Let radius of the circle be $r$.

So, ${(am-\alpha )}^2+{(\frac{a}{m}-\beta )}^2=r^2$

$\Rightarrow a^2{m}^2+{\alpha }^2-2m\alpha +\frac{a^2}{m^2}+{\beta }^2-\frac{2a\beta }{m}=r^2$

$\Rightarrow a^2{m}^4-2a{m}^3\alpha +m^2({\alpha }^2+{\beta }^2-r^2)-2a\beta m+a^2=0$

Since the above equation is biquadratic it has four roots.

$m_1,m_2,m_3$ are already roots of this equation.

Let fourth root be $m_4$

Parametric form of fourth point will be $(a{m}_4,\frac{a}{m_4})$

Product of roots i.e., $m_1{m}_2{m}_3{m}_4=\frac{a^2}{a^2}=1$

$\Rightarrow m_4=\frac{1}{m_1{m}_2{m}_3}$

i.e., Fourth point is $(\frac{a}{m_1{m}_2{m}_3},a{m}_1{m}_2{m}_3)$

Hence there exists a point $(\alpha ,\beta )$ which is equidistant from points $A,B,C,D$.