3let r≡(h,k) ,p≡(x1,y1) q≡(x2,y2)
y=3√3-x
put it in circle's equation we get x2=3√3 thus
x1+x2=0
centroid (a,b)=(h/3,6√3+k3)
thus h=3a , k=3b-6√3k
and it lies on circle given , so
put the value of h,k in x2+y2=27
we get radius as 3
Let PQ specified by the equation x + y = 3√3 be a chord to the circle x2+y2=27. If the point R moves on the circle, then the locus of centriod of triangle PQR is a circle, the square of whose radius is -
-
UP 0 DOWN 0 1 1
1 Answers
3