Question

Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first degree terms:
(i) y² + x² − 4x − 8y + 3 = 0
(ii) x² + y² − 5x + 2y − 5 = 0
(iii) x² − 12x + 4 = 0

Answer 1

Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.

(i) Substituting x = X + h and y = Y + k in the equation y² + x² − 4x − 8y + 3 = 0, we get:

$$\begin{gathered} {\left(Y+k\right)}^2+{\left(X+h\right)}^2-4\left(X+h\right)-8\left(Y+k\right)+3=0 \\ \Rightarrow Y^2+2kY+k^2+X^2+2hX+h^2-4X-4h-8Y-8k+3=0 \\ \Rightarrow X^2+Y^2+X\left(2h-4\right)+Y\left(2k-8\right)+k^2+h^2-4h-8k+3=0 \end{gathered}$$

For this equation to be free from the terms containing X and Y, we must have

$$\begin{gathered} 2h-4=0,2k-8=0 \\ \Rightarrow h=2,k=4 \end{gathered}$$

Hence, the origin should be shifted to the point (2, 4).

(ii) Substituting x = X + h and y = Y + k in the equation x² + y² − 5x + 2y − 5 = 0, we get:

$$\begin{gathered} {\left(X+h\right)}^2+{\left(Y+k\right)}^2-5\left(X+h\right)+2\left(Y+k\right)-5=0 \\ \Rightarrow X^2+2hX+h^2+Y^2+2kY+k^2-5X-5h+2Y+2k-5=0 \\ \Rightarrow X^2+Y^2+X\left(2h-5\right)+Y\left(2k+2\right)+k^2+h^2-5h+2k-5=0 \end{gathered}$$

For this equation to be free from the terms containing X and Y, we must have

$$\begin{gathered} 2h-5=0,2k+2=0 \\ \Rightarrow h=\frac{5}{2},k=-1 \end{gathered}$$

Hence, the origin should be shifted to the point $\left(\frac{5}{2},-1\right)$.

(iii) Substituting x = X + h and y = Y + k in the equation x² − 12x + 4 = 0, we get:

$$\begin{gathered} {\left(X+h\right)}^2-12\left(X+h\right)+4=0 \\ \Rightarrow X^2+2hX+h^2-12X-12h+4=0 \\ \Rightarrow X^2+X\left(2h-12\right)+h^2-12h+4=0 \end{gathered}$$

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 $\left(6,k\right),k\in R$.