Find the length of OA

k the esiest one :P

now |k-2h| =3√5 , |k-h|=3√2

solving this we get P.

then OA²=OP²-PA²

2x²+y²-3xy=0
i.e (y-x)(y-2x)=0
lines are y=x and y=2x

let C(a,b) be the center of the circle whose tangents are two lines y=x and y=2x

clearly b>a and b<2a

radius=3=length of perpendicular from C on the two tangents

=b-a/√2 = 2a-b/√5

from above, a=3(√5+√2) and b = 3(√5+2√2)

OA²=OC²- radius²

=(a²+b²-9
=9(7+2√10)+9(5+8+4√10)-9
=171+54√10=90+81+2√90x81=(√90+√81)²
therefroe OA=9+√90=9+3√10

Angle b/w tngnts = 2@

tan 2@ = 2√9/4 - 2 /(3) = 1/3
So, tan @ = √10 - 3

Also tan @ = 3/oa
So,oa = 3/(√10 - 3) = 3(√10 + 3)

wow.....three different methods..........[1]

let C(a,b) be the center of the circle whose tangents are two lines y=x and y=2xclearly b>a and b<2a

Plz explain this