Question
A triangle is formed by the lines whose combined equation is given by $$\left( {x + y - 4} \right)\left( {xy - 2x - y + 2} \right) = 0.$$ 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.
$${x^2} + {y^2} - 3x - 5y - 8 = 0$$
D.
none of these
Answer :
$${x^2} + {y^2} - 3x - 5y + 8 = 0$$
Solution :
Given combined equation is
$$\eqalign{
& \left( {x + y - 4} \right)\left( {xy - 2x - y + 2} \right) = 0 \cr
& \Rightarrow \left( {x + y - 4} \right)\left( {x - 1} \right)\left( {y - 2} \right) = 0 \cr} $$
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 $$A\left( {1,\,3} \right),\,B\left( {1,\,2} \right),\,C\left( {2,\,2} \right)$$
Slope of $$AC = \frac{{2 - 3}}{{2 - 1}} = - 1$$
So, slope of its perpendicular bisector is 1.
Mid-point of $$AC$$ is $$\left( {\frac{3}{2},\,\frac{5}{2}} \right)$$
So, equation of perpendicular bisector of $$AC$$ is
$$\eqalign{
& y - \frac{5}{2} = \left( {x - \frac{3}{2}} \right) \cr
& \Rightarrow 2y - 2x = 2 \cr
& \Rightarrow y - x = 1......\left( 1 \right) \cr} $$
Clearly, $$AB$$ makes an angle of $${90^ \circ }$$ with $$x$$-axis.
So, slope of perpendicular bisector of $$AB$$ is 0.
Mid-point of $$AB$$ is $$\left( {1,\,\frac{5}{2}} \right)$$
So, equation of perpendicular bisector of $$AB$$ is
$$y - \frac{5}{2} = 0......\left( 2 \right)$$
Solving equation (1) and (2), we get
$$x = \frac{3}{2},\,y = \frac{5}{2}$$
So, the coordinates of circumcenter is $$O\left( {\frac{3}{2},\,\frac{5}{2}} \right)$$
Radius $$ = OA = \frac{1}{{\sqrt 2 }}$$
Equation of circumcircle is
$$\eqalign{
& {\left( {x - \frac{3}{2}} \right)^2} + {\left( {y - \frac{5}{2}} \right)^2} = \frac{1}{2} \cr
& \Rightarrow 4{x^2} + 9 - 12x + 4{y^2} + 25 - 20y = 2 \cr
& \Rightarrow {x^2} + {y^2} - 3x - 5y + 8 = 0 \cr} $$