If \(\lambda\) is the perpendicular distance of a point \(\mathrm{P}\) on the circle \(x^2+y^2+2 x+2 y-3=0\), from the line \(2 x+y+13=0\), then maximum possible value of \(\lambda\) is

Given equation of the circle is \(x^2+y^2+2 x+2 y-3=0\)
Which can be written as: \((x+1)^2+(y+1)^2=5\) It is a circle with centre \((-1,-1)\) and radius \(\sqrt{5}\) Given line is: \(2 x+y+13=0\)
To find the required distance, we find the equation of a line perpendicular to the given line, and passing through the centre of the given circle.
\(\therefore\) Equation of this line is: \((y+1)=\frac{1}{2}(x+1)\) i.e., \(x=2 y+1\)
Now, we find the points where line \(x=2 y+1\) intersects the circle \(x^2+y^2+2 x+2 y-3=0\)
\(
\begin{array}{ll}
\therefore & (2 y+1)^2+y^2+2(2 y+1)+2 y-3=0 \\
\therefore & 4 y^2+4 y+1+y^2+4 y+2+2 y-3=0 \\
\therefore & 5 y^2+10 y=0 \\
\therefore & y(y+2)=0 \\
\therefore & y=0 \text { or } y=-2 \\
\therefore & x=1 \text { or } x=-3
\end{array}
\)
\(\therefore(1,0)\) and \((-3,-2)\) are the points on the circle, and one of them is at the maximum distance from the given line.
\(
\begin{array}{ll}
\therefore & \mathrm{d}_1=\left|\frac{2(1)+(0)+13}{\sqrt{4+1}}\right| \text { and } \mathrm{d}_2=\left|\frac{2(-3)+(-2)+13}{\sqrt{4+1}}\right| \\
\therefore & \mathrm{d}_1=\frac{15}{\sqrt{5}}=3 \sqrt{5} \quad \text { and } \mathrm{d}_2=\frac{5}{\sqrt{5}}=\sqrt{5} \\
\therefore & \lambda=3 \sqrt{5}
\end{array}
\)