Question

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

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

Answer 1

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

Let the sides of $△ABC$ be
$AB:x+y=6\cdots \left(i\right)BC:2x+y=4\cdots \left(ii\right)AC:x+2y=5\cdots \left(iii\right)$

The coordinates of $A,B$ and $C$ are
$A=(7,-1)B=(-2,8)C=(1,2)$

Let the equation of the circumcircle of $△ABC$ be
$x^2+y^2+2gx+2fy+c=0$
It passes through $A(7,-1),B(-2,8)$ and $C(1,2)$, so
$50+14g-2f+c=0\cdots \left(iv\right)68-4g+16f+c=0\cdots \left(v\right)5+2g+4f+c=0\cdots \left(vi\right)$

On solving these equations, we get
$g=-\frac{17}{2},f=-\frac{19}{2}$ and $c=50$

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