The equation ^2 = ( x + y )^24xy is only possible when ?

Answer: Option C

$$\begin{array}{rl}& {\cos }^2\theta =\frac{{\left(x+y\right)}^2}{4xy}\\ & \text{Max value of }{\cos }^2\theta =1\\ & \Rightarrow 1=\frac{{\left(x+y\right)}^2}{4xy}\\ & \Rightarrow 4xy={\left(x+y\right)}^2\\ & \Rightarrow 4xy=x^2+y^2+2xy\\ & \Rightarrow 0=x^2+y^2-2xy\\ & \Rightarrow 0={\left(x-y\right)}^2\\ & \Rightarrow 0=x-y\\ & \Rightarrow x=y\end{array}$$