Question 631573
(1+m^2)x^2+2mcx+(c^2-a^2)=0

Since the quadratic equation has equal roots the discriminant 
{{{b^2-4ac=0}}}

where {{{a= (1+m^2)}}}   , b= 2mc , {{{c=(c^2-a^2)}}}

{{{(2mc)^2- 4(1+m^2)(c^2-a^2)=0}}}

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

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

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

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

divide by 4
{{{c^2=a^2+a^2m^2}}}

{{{c^2=a^2(1+m^2)}}}

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