gangula deekshith
Last Activity: 2 Years ago
given circle x2+y2=4
then the radius of the circle is r=2
we know that the circle centre (g,f) lies on the origin (0,0)
given that PAB is equilateral triangle so ang(A)=ang(B)=ang(P)=600
let PD be the altittude of the triangle then ang(OPB)=ang(OPA)=300
according to pythagorus theorem
consider triangleOPB
here OB=2 (OB is the raduis of the circle)
cos300=BP/OA
3½/2=BP/2
BP=3½
similarly AP=31/2
AP+BP=2(31/2)
we know that (x-h)2+(y-k)2=0 here (h,k)=(31/2,31/2)
x2+2(31/2)x -8 +y2+2(31/2)y-8=0
x2+y2-16=0 [cirlce centre lies on the origin so (g,f)=0
x2+y2=16
