Question

Without changing the direction of coordinates axes, origin is transferred to $(\alpha ,\beta )$ so that linear term in the equation $x^2+y^2+2x-4y+6=0$ are eliminated the point $(\alpha ,\beta )$ is

• A. $(-1,2)$
• B. $(1,-2)$
• C. $(1,2)$
• D. $(-1,-2)$

Answer 1

The correct option is A $(-1,2)$

$x=x^{\prime }+\alpha ,y=y^{\prime }+\beta$

Substitute $x$, y in given equ',

${(x^{\prime }+\alpha )}^2+{(y^{\prime }+\beta )}^2+2(x^{\prime }+\alpha )-4(y^{\prime }+\beta )+6=0$

$\begin{array}{cc}\Rightarrow & x^{\prime 2}+y^{\prime 2}+x^{\prime }(2\alpha +2)+y^{\prime }(2\beta -4)+6+{\alpha }^2\\ & +{\beta }^2+2\alpha -4\beta =0\\ \Rightarrow & 2\alpha +2=0,2\beta -4=0\end{array}$

$\alpha =\frac{-2}{2}=-1$ (to eliminate linear terns)

$\beta =\frac{4}{2}=2$

Option A is correct.