Q. Find the equation of the circle circumscribing the triangle formed by the lines :x+y=6 , 2x+y=4, and x+2y=5 , without finding the vertices of the triangle?
11This one is not "without finding the vertices of the triangle" ..... but a gud one (thnx to tapan.)
Vertices : (-2,8),(7,-1),(1,2)
Eqn. of circumcircle : (x-1)(x-7) + (y-2)(y+1) + λ(x+2y-5) = 0
Substituting (-2,8) in the eqn. we get λ = -9
Hence, Eqn of circumcircle : (x-1)(x-7) + (y-2)(y+1) - 9(x+2y-5) = 0
11its a question given in a subjective test
it was mentioned "without finding the vertices of the triangle"
i wrote it to search for a new method that i am missing.......
1very easy..equation of circumcircle is L1L2+L2L3+L3L1+λ=0...
The value of λ can be found out using the characteristics of the second degree equation of a circle(i.e. coeffx=coeffy ....& g2+f2-c>0)
1L1,L2 & L3 are the 3 sides of a triangle...
using
LiL2+L2L3+L3L1+λ=0 u will get...sumthing like ths
ax2
+by2+2gx+2fy+k=o
for this to be a circle...a=b and
g2
+f2-k>0
GOT IT NOW??
1Eqn is L1L2+λL2L3+μL3L1=0
provided coefficient of xy=0
coefficient of x2= coefficient ofy2
