1. Tardigrade
  2. Question
  3. Mathematics
  4. The equations of the sides BC , CA and AB of a triangle A B C are x+y=4,2 x+y=3 and x+2 y=3 respectively. Let S1, S2 and S3 are the circle drawn on BC , C A 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 :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The equations of the sides $BC , CA$ and $AB$ of a triangle $A B C$ are $x+y=4,2 x+y=3$ and $x+2 y=3$ respectively. Let $S_{1}, S_{2}$ and $S_{3}$ are the circle drawn on $BC , C A$ 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 :

Conic Sections

Solution:

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