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 circumcircle is

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

Answer 1

Answer: (B)

$x^2+y^2-3x-5y+8=0$

Given combined eqn is
$(x+y-4)(xy-2x-y+2)=0$
$\Rightarrow (x+y-4)(x-1)(y-2)=0$
So, the equation of sides of the triangle are $x=1,y=2,x+y=4$
Point of intersection of any two sides of a triangle will give the vertices of triangle .
So, the vertices of the triangle are $A(1,3),B(1,2),C(2,2)$
Slope of $AC$ $=\frac{2-3}{2-1}=-1$
So, slope of its perpendicular bisector is $1$.
Mid-point of $AC$ is $(\frac{3}{2},\frac{5}{2})$.
So, equation of perpendicular bisector of $AC$ is
$y-\frac{5}{2}=(x-\frac{3}{2})$
$\Rightarrow 2y-2x=2$
$\Rightarrow y-x=1$ ....(1)
Clearly, $AB$ makes an angle of ${90}^0$ with $x$-axis.
So, slope of perpendicular bisector of AB is $0$.
Mid-point of $AB$ is $(1,\frac{5}{2})$
So, eqn of perpendicular bisector of AB is
$y-\frac{5}{2}=0$ .....(2)
Solving (1) and (2), we get
$x=\frac{3}{2},y=\frac{5}{2}$
So, the coordinates of circumcenter is $O(\frac{3}{2},\frac{5}{2})$.
Radius $=OA=\frac{1}{\sqrt{2}}$
Equation of circumcircle is
${(x-\frac{3}{2})}^2+{(y-\frac{5}{2})}^2=\frac{1}{2}$
$\Rightarrow 4x^2+9-12x+4y^2+25-20y=2$
$\Rightarrow x^2+y^2-3x-5y+8=0$