Question

The equation of the circle circumscribing the triangle formed by the lines $x+y=6,2x+y=4$ and $x+2y=5$ is

• A. $x^2+y^2-17x-19y+50=0$
• B. $x^2+y^2+17x+19y+50=0$
• C. $x^2+y^2-19x-17y+50=0$
• D. $x^2+y^2+19x-17y+50=0$

Answer 1

The correct option is A $x^2+y^2-17x-19y+50=0$

Given lines
$L_1:x+y-6=0L_2:2x+y-4=0L_3:x+2y-5=0$
Consider the equation
$L_1{L}_2+\lambda L_2{L}_3+\mu L_3{L}_1=0$
$\Rightarrow (x+y-6)(2x+y-4)+\lambda (2x+y-4)(x+2y-5)+\mu (x+2y-5)(x+y-6)=0\cdots \left(i\right)$
Where $\lambda$ and $\mu$ are constants.

This is a circle, when
coefficient of $x^2=$ coefficient of $y^2$
$\Rightarrow 2+2\lambda +\mu =1+2\lambda +2\mu \Rightarrow \mu =1$

Coefficient of $xy=0$
$\Rightarrow 3+5\lambda +3\mu =0\Rightarrow \lambda =-\frac{6}{5}$

On substituting the values of $\lambda$ and $\mu$ in equation $\left(i\right)$, we get
$L_1{L}_2-\frac{6}{5}{L}_2{L}_3+L_3{L}_1=0\Rightarrow 5L_1[L_2+L_3]=6L_2{L}_3\Rightarrow 5(x+y-6)3(x+y-3)=6(2x+y-4)(x+2y-5)\Rightarrow 5[x^2+y^2+2xy-9x-9y+18]=2[2x^2+2y^2+5xy-14x-13y+20]\Rightarrow x^2+y^2-17x-19y+50=0$


Alternate Solution:
Let the equation of side be
$AB:x+y=6BC:2x+y=4AC:x+2y=5$

Finding the point of intersection, we get the coordinates of the vertices as
$A=(7,-1),B=(-2,8),C=(1,2)$

Let the equation of the circumcircle of $△ABC$ be
$x^2+y^2+2gx+2fy+c=0\cdots \left(ii\right)$

It passes through $A(7,-1),B(-2,8)$ and $C(1,2)$, so
$50+14g-2f+c=068-4g+16f+c=05+2g+4f+6c=0$
On solving these equations, we get
$g=-\frac{17}{2},f=-\frac{19}{2},c=50$

Hence, the equation of the circumcircle is
$x^2+y^2-17x-19y+50=0$