Question

The values of $x$ and $y$ such that $y$ satisfy the equation $(x,y\in$ real numbers) $x^2-xy+y^2-4x-4y+16=0$ is

• A. $4,4$
• B. $3,3$
• C. $2,2$
• D. none of these

Answer 1

Answer: (A)

Given, $x^2-xy+y^2-4x-4y+16=0$
$\Rightarrow$ $x^2-(y+4)x+y^2-4y+16=0$
For real $x,{(y+4)}^2-4(y^2-4y+16)\ge 0$
$\Rightarrow$ $-3y^2+24y-48=0$
$\Rightarrow$ $y^2-8y+16=0$
$\Rightarrow$ ${(y-4)}^2=0\Rightarrow y=4$
$\therefore$ $x=4$ $\Rightarrow$ $(x,y)=(4,4)$