Answer 1

Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.
Substituting x = X + h and y = Y + k in the equation x² − 12x + 4 = 0, we get:
\[\left( X + h \right)^2 - 12\left( X + h \right) + 4 = 0\]
\[ \Rightarrow X^2 + 2hX + h^2 - 12X - 12h + 4 = 0\]
\[ \Rightarrow X^2 + X\left( 2h - 12 \right) + h^2 - 12h + 4 = 0\]
For this equation to be free from the terms containing X and Y, we must have

\[2h - 12 \Rightarrow h = 6\]
Hence, the origin should be shifted to the point \[\left( 6, k \right), k \in R\].