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 $x^2+y^2-17x-19y+50=0$

Let the equation of side $AB,BC$ and $CA$ of $△ABC$ are respectively
$x+y=6$,
$2x+y=4$
and $x+2y=5$

The coordinates of $A,B$ and $C$ are $(7,-1),(-2,8)$ and $(1,2)$ respectively.
Let the equation of the circumcircle of $△ABC$ be
$x^2+y^2+2gx+2fy+c=0\dots \left(i\right)$
It passes through $A(7,-1),B(-2,8)$ and $C(1,2)$
Therefore,
$50+14g-2f+c=0$
$68-4g+16f+c=0$
and $5+2g+4f+6c=0$
On solving these equations, we get
$g=-\frac{17}{2},f=-\frac{19}{2}$ and $c=50$
On substituting the value of $g,f$ and $c$ in Equation $\left(i\right)$, the equation of the required circumcircle is
$x^2+y^2-17x-19y+50=0$

Aliternative
Consider the equation
$(x+y-6)(2x+y-4)+\lambda (2x+y-4)(x+2y-5)+\mu (x+2y-5)(x+y-6)=0$
where $\lambda$ and $\mu$ are constants.
This equation represents a curve passing through the vertices of the triangle formed by the given lines.
We have to determine the values of $\lambda$ and $\mu$ so that the curve given in Equation $\left(i\right)$ is a circle.
The curve given in Equation $\left(i\right)$ will be a circle, if
coefficient of $x^2$ = coefficient of $y^2$
and coefficient of $xy=0$
$\Rightarrow 2+2\lambda +\mu =1+2\lambda +2\mu 3+5\lambda +3\mu =0$
$\Rightarrow \mu =1$ and $3\mu +5\lambda +3=0$
$\Rightarrow \mu =1and\lambda =\frac{-6}{5}$
On substituting the values of $\lambda$ and $\mu$ in Eq. $\left(i\right)$, we get
$(x+y-6)(2x+y-4)-\left(\frac{6}{5}\right)(2x+y-4)(x+2y-5)+(x+2y-5)(x+y-6)=0$
$\Rightarrow x^2+y^2-17x-19y+50=0$