A rod of length 8 units having two end points always lie on \(x-y+2=0\) and \(x+y+2=0\). A point \(P\) divide this line …

\(\begin{aligned}
& x-y+2=0 \\
& x+y+2=0
\end{aligned}\)

\( P Q=8 \)
\(\Rightarrow(a-b)^2+(a+2+b+2)^2=64 \)
\((a-b)^2+(a+b+4)=64\)
\(2 a^2+2 b^2-2 a b+16+2(a b+4 b+4 a)=64 \)
\(\Rightarrow 2 a^2+2 b^2+8 a+8 b=48 \)
\(\Rightarrow a^2+b^2+4 a+4 b-24=0\)
\(\Rightarrow (a+2)^2+(b+2)^2=32\)
\begin{aligned}
& h=\frac{2 b+1(a)}{3}, k=\frac{2(-2-b)+1(a+2)}{3} \\
& 3 h=2 b+a, 3 k=a-2-2 b \rightarrow \text { Solve for } a \text { and } b . \\
& a=\frac{3 h+3 k+2}{2}, b=\frac{3 h-3 k-2}{4} \\
& (a+2)=\left(\frac{3 h+3 k+6}{2}\right), b+2=\left(\frac{3 h-3 k+6}{2}\right) \\
& \Rightarrow\left(\frac{3 h+3 k+6}{2}\right)^2+\left(\frac{3 h-3 k+6}{2}\right)^2=32 \\
& 9(x+y+2)^2+9(x-y+2)^2=128 \\
& 18\left[x^2+y^2+4 x+4\right]=128 \Rightarrow x^2+y^2+4 x-\frac{28}{9}=0
\end{aligned}\)