The equation of the circle with (1, 1) as centre and which cuts a chord of length $4\sqrt{2}$ units on the line $x+y+1=0$is

$$\begin{gathered} \mathrm{Given} \mathrm{line} \mathrm{x}+\mathrm{y}+1=0 \\ \mathrm{and} \mathrm{centre}\left(\mathrm{C}\right)=(1, 1) \\ Now \mathrm{d}=perpendicular dis\tan ce from (1,1) to \left(1\right) \\ =\frac{\left|3\right|}{\sqrt{2}} \end{gathered}$$

Given Length of chord $=4\sqrt{2}$

$$\begin{gathered} 4\sqrt{2}=2\sqrt{{\mathrm{r}}^2-{\mathrm{d}}^2} \\ 16\left(2\right)=4\left({\mathrm{r}}^2-\frac{9}{2}\right) \\ \Rightarrow 8={\mathrm{r}}^2-\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \Rightarrow {\mathrm{r}}^2=8+\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=\frac{25}{2} \\ \Rightarrow \mathrm{r}=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right. \\ {So,circle is (\mathrm{x}-1)}^2+{(\mathrm{y}-1)}^2=\frac{25}{2} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-2\mathrm{y}-\frac{21}{2} =0 \\ \Rightarrow 2{\mathrm{x}}^2+2{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}-21=0 \end{gathered}$$