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Quadratic equation$:(a^{2}+b^{2})x^{2}+2(b^{2}+c^{2})x+(b^{2}+c^{2})= 0\rightarrow(1)$

For any quadratic equation written in standard form of $ax^2+bx+c=0$. The discriminant $D$ for the quadratic equation is

$$D=b^2-4ac,$$

where

$$\begin{cases} b^2-4ac \gt 0: & \text{two distinct real roots} \\ b^2-4ac=0: & \text{equal and real roots} \\ b^2-4ac \lt 0: & \text{imaginary roots}. \end{cases}$$

Now, from the equation $(1),$ we get

Roots are real $:b^{2} – 4ac\geq 0$

$\implies[2(b^{2} + c^{2})]^{2} – 4(a^{2} + b^{2})(b^{2} + c^{2}) \geq 0$

$\implies 4 (b^{2} + c^{2})^{2} – 4(a^{2} + b^{2})(b^{2} + c^{2}) \geq 0$

$\implies 4 (b^{2} + c^{2})[(b^{2} + c^{2}) – (a^{2} + b^{2})]\geq 0$

$\implies 4 (b^{2} + c^{2})[b^{2} + c^{2} – a^{2} - b^{2}]\geq 0$

$\implies 4 (b^{2} + c^{2})(c^{2} – a^{2})\geq 0$

We can write like $4 (b^{2} + c^{2}) \geq 0 \:\&(c^{2} – a^{2})\geq 0$

$\implies c^{2}\geq a^{2}$

So, the correct answer is $(A).$