Question

A triangle is formed by the lines whose combined equation is given by $\left(x+y-4\right)\left(xy-x-y+2\right)=0$ 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 1

The correct option is B

$x^2+y^2–3x–5y+8=0$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\left(x+y-4\right)\left(xy-x-y+2\right)=0$

$\Rightarrow$ $\left(x+y-4\right)\left(x-1\right)\left(y-2\right)=0$

Therefore, the sides of triangle are $x+y-4=0,x-1=0\&y-2=0$.

Step-2: Draw the triangle using given sides.

We know that In a right angle triangle circumcentre is on mid-point of hypotenuse.

Therefore, circumcentre is $\left(\frac{1+2}{2},\frac{3+2}{2}\right)$.

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

Step-3: Equation of circumcircle.

we know that equation of circle having center $\left(x_1,y_1\right)$ and radius $r_1$ is ${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2={r_1}^2$

Therefore equation of circumcircle is ${\left(x-\frac{3}{2}\right)}^2+{\left(y-\frac{5}{2}\right)}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$

$\Rightarrow$ $x^2+\frac{9}{4}-3x+y^2+\frac{25}{4}-5y=\frac{1}{2}$

$\Rightarrow$ $\frac{4x^2-12x+4y^2-20y+34-2}{4}=0$

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

$\Rightarrow$ $\left(x^2+y^2-3x-5y+8\right)=0$

Hence correct answer is option $\left(B\right)$