The sum of infinite terms of a G.P. x and on squaring each term of it, the will be y, then find the common ratio of this series.
A.\[\dfrac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}\]
B. \[\dfrac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}\]
C. \[\dfrac{{{x^2} - y}}{{{x^2} + y}}\]
D. \[\dfrac{{{x^2} + y}}{{{x^2} - y}}\]

Hints First write the sum formula of infinite terms of a G.P. then equate that with x and obtain the value of a from there. Now, write the sum formula of an infinite G.P series with square terms and substitute the value of a and calculate to obtain the required result.

Formula used
The sum formula of an infinite G.P series \[a,ar,a{r^2},a{r^3},....\] is
\[S = \dfrac{a}{{1 - r}}\] , where a is the first term and r is the common ratio.

Complete step by step solution
The sum formula of an infinite G.P series \[a,ar,a{r^2},a{r^3},....\] is
\[S = \dfrac{a}{{1 - r}}\] .
It is given that,
\[x = \dfrac{a}{{1 - r}}\]
\[a = x(1 - r)\]
Now, the sum formula of an infinite G.P series \[{a^2},{a^2}{r^2},{a^2}{r^4},{a^2}{r^6},....\] is
\[S = \dfrac{{{a^2}}}{{1 - {r^2}}}\] .
It is given that,
\[y = \dfrac{{{a^2}}}{{1 - {r^2}}}\]
Substitute \[a = x(1 - r)\] in the equation \[y = \dfrac{{{a^2}}}{{1 - {r^2}}}\] for further calculation.
\[y = \dfrac{{{{\left[ {x(1 - r)} \right]}^2}}}{{1 - {r^2}}}\]
\[y = \dfrac{{{x^2}{{(1 - r)}^2}}}{{(1 - r)(1 + r)}}\]
\[\dfrac{y}{{{x^2}}} = \dfrac{{1 - r}}{{1 + r}}\]
\[\dfrac{{{x^2}}}{y} = \dfrac{{1 + r}}{{1 - r}}\]
Apply componendo dividendo rule,
\[\dfrac{{{x^2} - y}}{{{x^2} + y}} = \dfrac{{\left( {1 + r} \right) - (1 - r)}}{{\left( {1 + r} \right) + (1 - r)}}\]
\[\dfrac{{{x^2} - y}}{{{x^2} + y}} = \dfrac{{2r}}{2}\]
\[r = \dfrac{{{x^2} - y}}{{{x^2} + y}}\]
The correct option is C.

Note Here we have obtained the common ratio of the GP. Always remember that the common ratio in a G.P series is always constant. One should not get confused with the G.P and A.P.