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(A) Given $|\overrightarrow{x}+\overrightarrow{y}|=|\overrightarrow{x}|$. Squaring both sides,we get:$|\overrightarrow{x}|^2+|\overrightarrow{y}|^2+2\

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(A) Given $|\overrightarrow{x}+\overrightarrow{y}|=|\overrightarrow{x}|$. Squaring both sides,we get:$|\overrightarrow{x}|^2+|\overrightarrow{y}|^2+2\overrightarrow{x}\cdot\overrightarrow{y}=|\overrightarrow{x}|^2$$|\overrightarrow{y}|^2+2\overrightarrow{x}\cdot\overrightarrow{y}=0$  --- $(1)$Given that $(2\overrightarrow{x}+\lambda\overrightarrow{y})$ is perpendicular to $\overrightarrow{y}$,their dot product is zero:$(2\overrightarrow{x}+\lambda\overrightarrow{y})\cdot\overrightarrow{y}=0$$2\overrightarrow{x}\cdot\overrightarrow{y}+\lambda|\overrightarrow{y}|^2=0$  --- $(2)$From equation $(1)$,we have $2\overrightarrow{x}\cdot\overrightarrow{y}=-|\overrightarrow{y}|^2$. Substituting this into equation $(2)$:$-|\overrightarrow{y}|^2+\lambda|\overrightarrow{y}|^2=0$$(\lambda-1)|\overrightarrow{y}|^2=0$Since $\overrightarrow{y}$ is a non-zero vector,$|\overrightarrow{y}|^2 
eq 0$. Therefore,$\lambda-1=0$,which gives $\lambda=1$.

If $\overrightarrow{x}$ and $\overrightarrow{y}$ are two non-zero vectors such that $|\overrightarrow{x}+\overrightarrow{y}|=|\overrightarrow{x}|$ and $2\overrightarrow{x}+\lambda\overrightarrow{y}$ is perpendicular to $\overrightarrow{y}$,then the value of $\lambda$ is

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