Question

A triangle is formed by the lines whose combined equation is given by $(x+y-4)(xy-2x-y+2)=0$. The equation of its circum-circle is

• A. $x^2+y^2-5x-3y+8=0$
• B. $x^2+y^2-3x-5y+8=0$
• C. $x^2+y^2+2x+2y-3=0$
• D. None of these

Answer 1

The correct option is B $x^2+y^2-3x-5y+8=0$

Given, $(x+y-4)\left((x-1)(y-2)\right)=0$
Solving $x-1=0,x+y-4$ we get $x=1,y=3.$
Solving $y-2=0,x+y-4=0$
we get $x=2,y=2.$
They are and points of diameter $(x-2)(x-1)+(y-2)(y-3)=0$
$\Rightarrow x^2+y^2-3x-5y+8=0$