If the equation (1+m2)x2+2 mcx+(c2−a2)=0 has equal roots,

The given equation $\left(1+m^{2}\right) x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0$, has equal roots

Then prove that $c^{2}=\left(1+m^{2}\right)$.

Here,

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

As we know that $D=b^{2}-4 a c$

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

$D=b^{2}-4 a c$

$=\{2 m c\}^{2}-4 \times\left(1+m^{2}\right) \times\left(c^{2}-a^{2}\right)$

$=4\left(m^{2} c^{2}\right)-4\left(c^{2}-a^{2}+m^{2} c^{2}-m^{2} a^{2}\right)$

$=4 a^{2}+4 m^{2} a^{2}-4 c^{2}$

The given equation will have real roots, if $D=0$

$4 a^{2}+4 m^{2} a^{2}-4 c^{2}=0$

$4 a^{2}+4 m^{2} a^{2}=4 c^{2}$

$4 a^{2}\left(1+m^{2}\right)=4 c^{2}$

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

Hence,

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