If the equations $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+c^{2}+d^{2}=0$ has equal roots, then
(a) $a b=c d$
(b) $a d=b c$
(c) $a d=\sqrt{b c}$
(d) $a b=\sqrt{c d}$
The given quadric equation is $\left(a^{2}+b^{2}\right) x^{2}-2(a c+b d) x+c^{2}+d^{2}=0$, and roots are equal.
Here, $a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=c^{2}+d^{2}$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=\left(a^{2}+b^{2}\right), b=-2(a c+b d)$ and, $c=c^{2}+d^{2}$
$=\{-2(a c+b d)\}^{2}-4 \times\left(a^{2}+b^{2}\right) \times\left(c^{2}+d^{2}\right)$
$=4 a^{2} c^{2}+4 b^{2} d^{2}+8 a b c d-4\left(a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}\right)$
$=+8 a b c d-4 a^{2} d^{2}-4 b^{2} c^{2}$
$=-4\left(a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right)$
The given equation will have equal roots, if $D=0$
$-4\left(a^{2} d^{2}+b^{2} c^{2}-2 a b c d\right)=0$
$a^{2} d^{2}+b^{2} c^{2}-2 a b c d=0$
$(a d-b c)^{2}=0$
$a d-b c=0$
$a d=b c$
Thus, the correct answer is (b)