$\text{ For }a,b,c\text{ non-zero, real distinct, the equation, }\left(a^2+b^2\right)x^2-2b(a+c)x+b^2+c^2=0\text{ has non-zero real } \mathrm{roots}.\mathrm{One} \mathrm{of} \mathrm{these} roots$$\text{ is also the root of the equation: }$

$\begin{array}{l}\left(a^2+b^2\right)x^2-2b(a+c)x+b^2+c^2=0\\ D=4b^2(a+c{)}^2-4\left(a^2+b^2\right)\left(b^2+c^2\right)\\ =-4\left(b^4-2b^{2}ac+a^2{c}^2\right)\end{array}$

$=-4{\left(b^2-ac\right)}^2$

$\text{ For real roots, }D\ge 0$

$\begin{array}{l}\Rightarrow -4{\left(b^2-ac\right)}^2\ge 0\\ \Rightarrow b^2-ac=0\\ \Rightarrow \text{ Roots are real and equal. }\end{array}$

$\begin{array}{l}\therefore \text{ Roots are }\frac{2b(a+c)\pm 0}{2\left(a^2+b^2\right)}\\ =\frac{b(a+c)}{a^2+ac}\\ =\frac{b}{a}\end{array}$

This root satisfies option (c).