Question

${(x-1)}^2+{(y-2)}^2=3{\left(\frac{x+y-3}{\sqrt{2}}\right)}^2$

Answer 1

Consider the given equation,

${(x-1)}^2+{(y-2)}^2=3{\left(\frac{x+y-3}{\sqrt{2}}\right)}^2$

$x^2+1^2-2x+y^2+2^2-2y=\frac{3{(x+y-3)}^2}{2}$

$x^2+1-2x+y^2+4-2y=\frac{3(x^2+y^2+{(-3)}^2+2xy+2y(-3)+2(-3)x)}{2}$

$x^2+y^2-2x-2y+5=\frac{3(x^2+y^2+9+2xy-6y-6x)}{2}$

$\frac{1}{2}{x}^2+\frac{1}{2}{y}^2+3xy-7x-7y+\frac{17}{2}=0$

Hence, this is the answer,