$\text{ If }\mathrm{y}=\mathrm{mx}+\mathrm{c}\text{ and }{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2\text{ intersects at }\mathrm{A}\text{ and }\mathrm{B}\text{, if }\mathrm{AB}=2\mathrm{\lambda }\text{, then show that }{\mathrm{C}}^2=\left(1+{\mathrm{m}}^2\right)\left({\mathrm{a}}^2-{\mathrm{\lambda }}^2\right)$

$\text{ Given circle is }{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$
Center O = (0,0) ; Radius r = a
Equation of chord AB is y = mx + c
$\Rightarrow \mathrm{mx}-\mathrm{y}+\mathrm{c}=0$
$\mathrm{d}={\perp }^{\text{ar }}\text{ distance from center ' }\mathrm{O}\text{ ' to chord }$
$\mathrm{AB}=\frac{\left|\mathrm{C}\right|}{\sqrt{{\mathrm{m}}^2+1}}$
Length of chord $\mathrm{AB}=2\mathrm{\lambda }\text{ (given) }$

$\Rightarrow 2\sqrt{{\mathrm{r}}^2-{\mathrm{d}}^2}=2\mathrm{\lambda }$
$\left[\therefore \text{ Length of the chord }\mathrm{AB}=2\sqrt{{\mathrm{r}}^2-{\mathrm{d}}^2}\right]$
$\begin{array}{l}\Rightarrow {\mathrm{r}}^2-{\mathrm{d}}^2={\mathrm{\lambda }}^2\Rightarrow {\mathrm{r}}^2-{\mathrm{\lambda }}^2={\mathrm{d}}^2\\ \Rightarrow {\mathrm{a}}^2-{\mathrm{\lambda }}^2=\frac{{\mathrm{c}}^2}{1+{\mathrm{m}}^2}\Rightarrow {\mathrm{c}}^2=\left(1+{\mathrm{m}}^2\right)\left({\mathrm{a}}^2-{\mathrm{\lambda }}^2\right)\end{array}$