Let the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}+6\mathrm{y}+\mathrm{\lambda }=0(\mathrm{\lambda }\in \mathrm{R})$bisect the circumference of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+2\mathrm{y}+(\cos \mathrm{\theta }+\sin \mathrm{\theta })=0,(\mathrm{\theta }\in \mathrm{R})$ Then the product of maximum and minimum values of $\mathrm{\lambda }$ is

Equation of common chord is $6\mathrm{x}+4\mathrm{y}+\mathrm{\lambda }-\cos \mathrm{\theta }-\sin \mathrm{\theta }=0$
It passes through (1,-1).
$\begin{array}{l}\therefore 6-4+\mathrm{\lambda }=\cos \mathrm{\theta }+\sin \mathrm{\theta }\\ \Rightarrow \mathrm{\lambda }=\cos \mathrm{\theta }+\sin \mathrm{\theta }-2\\ \Rightarrow {\mathrm{\lambda }}_{max}=\sqrt{2}-2\text{ and }{\mathrm{\lambda }}_{min}=-\sqrt{2}-2\\ \Rightarrow \text{ Product }=(2-\sqrt{2})(2+\sqrt{2})=2\end{array}$