Question

The equations of the sides $BC,CA$ and $AB$ of a triangle $ABC$ are $x+y=4,2x+y=3$ and $x+2y=3$ respectively. Let $S_1,S_2$ and $S_3$ are the circle drawn on $BC,CA$ and $\Lambda B$ as diameters respectively. The radical axes of the circles $S_1$ and $S_2;S_2$ and $S_3;S_3$ and $S_1$ are represented by the lines $L_{12},L_{23}$ and $L_{31}$ respectively, which intersect the sides $AB,BC$ and CA at respectively.
Equation of the circumcircle of triangle $AQC$ is :

• A. $x^2+y^2-4x-4y-10=0$
• B. $x^2+y^2-6y+4=0$
• C. $x^2+y^2-6x+4-0$
• D. $x^2+y^2+14x+14y-46=0$

Answer 1

Answer: (B)

The equations of the sides are
$BC:x+y-4=0$....(i)
$CA:2x+y-3=0$ .... (ii)
$AB:x+2y-3=0$ .... (iii)
Solving these three lines, we get $A(1,1),B(5,-1)$ and $C(-1,5)$
The equation of the circles are
$S_1:x^2+y^2-4x-4y-10=0$ ...(iv)
$S_2:x^2+y^2-6y+4=0$ .... (v)
$S_3:x^2+y^2-6x+4=0$... (vi)
The radical axes are also the altitudes of the triangle. Thus $AQ$ is perpendicular to $BC$. Thus, the circle $S2$ with $AC$ as diameter passes through the point $Q$. Hence, $S_2$ is the circle circumscribing the triangle $AQC$