Question

If point $P_n=(\log a_n,\frac{1}{\log a_n}),n=1,2,3,4$, where $a_n>0,a_n\ne 1$ lies on the circle whose centre is on $y-$axis, then the value of $a_1{a}_2{a}_3{a}_4$ is

Answer 1

Let the circle whose centre is on $y-$axis be
$x^2+(y-b{)}^2=r^2$
$\Rightarrow x^2+y^2-2yb+b^2-r^2=0\cdots \left(1\right)$
Given that $P_n=(\log a_n,\frac{1}{\log a_n})$ lies on the circle
Therefore,
$(\log a_n{)}^2+\frac{1}{(\log a_n{)}^2}-\frac{2b}{\log a_n}+b^2-r^2=0\Rightarrow (\log a_n{)}^4+1-2b\log a_n+(b^2-r^2)(\log a_n{)}^2=0$
Whose roots are $\log a_1,\log a_2,\log a_3,\log a_4$
Assuming $\lambda =\log a_n$
$\Rightarrow {\lambda }^4+(b^2-r^2){\lambda }^2-2b\lambda +1=0$
So, the sum of the roots = 0
$\log a_1+\log a_2+\log a_3+\log a_4=0$
$\Rightarrow \log \left(a_1{a}_2{a}_3{a}_4\right)=0\therefore a_1{a}_2{a}_3{a}_3=1$