$\left(1+m^{2}\right) x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0$

Here,

$a=\left(1+m^{2}\right), b=2 m c \text { and } c=\left(c^{2}-a^{2}\right)$

It is given that the roots of the equation are equal; therefore, we have:

$\begin{array}{l}
D=0 \\
\Rightarrow\left(b^{2}-4 a c\right)=0 \\
\Rightarrow(2 m c)^{2}-4 \times\left(1+m^{2}\right) \times\left(c^{2}-a^{2}\right)=0
\end{array}$

$\begin{array}{l}
\Rightarrow 4 m^{2} c^{2}-4\left(c^{2}-a^{2}+m^{2} c^{2}-m^{2} a^{2}\right)=0 \\
\Rightarrow 4 m^{2} c^{2}-4 c^{2}+4 a^{2}-4 m^{2} c^{2}+4 m^{2} a^{2}=0 \\
\Rightarrow-4 c^{2}+4 a^{2}+4 m^{2} a^{2}=0 \\
\Rightarrow a^{2}+m^{2} a^{2}=c^{2} \\
\Rightarrow a^{2}\left(1+m^{2}\right)=c^{2} \\
\Rightarrow c^{2}=a^{2}\left(1+m^{2}\right)
\end{array}$

Hence proved.