circles(2)

This 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

its 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.......

very 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)

kamal....pl in detail...

L1,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??

wah gr8888 tx..:-)

Eqn is L1L2+λL2L3+μL3L1=0
provided coefficient of xy=0
coefficient of x2= coefficient ofy2

jeepers...even i 4got dat1[5]