Question

The equation $(x+y-6)(xy-3x-y+3)=0$ represent the sides of a triangle then the equation of the circumcircle of the triangle is

• A. $x^2+y^2-5x-9y+20=0$
• B. $x^2+y^2-4x-8y+18=0$
• C. $x^2+y^2-3x-5y+8=0$
• D. $x^2+y^2+2x-3y-1=0$

Answer 1

Answer: (B)

$x^2+y^2-4x-8y+18=0$

We have $x+y=6$ and $xy-y-3x+3=0$

$\Rightarrow y(x-1)-3(x-1)=0\Rightarrow (x-1)(y-3)=0$

Equation of the sides of the triangle are

$x+y=6$ ...(1)

$y=3$ ...(2)

$x=1$ ...(3)

Shaded triangle is right angled at $(1,3)$

$\therefore$ the circumcircle is the circle on $(3,3)$ and $(1,5)$ as ends of a diameter and its equation is

$(x-3)(x-1)+(y-3)(y-5)=0\Rightarrow x^2+y^2-4x-8y+18=0$