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 …

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

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