19i was in for coordinate geometry...was solving the problem nicely !
Prove that every circle passing through the points z_{0} and 1/z 0 intersects the circle |z|=1 at right angles
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8 Answers
62take the equation of the circle which passes through
(r cos θ, r sin θ)
and (1/r cos θ, 1/r sin θ)
I havent tried it but i guess you should solve it by coordinate geometry or even give pure geometry a change..
i am getting late otherwise would have tried this one for sure...
19
341hint: what is the length of the tangent drawn from origin to the given circle?
3general eqxn of circle passin thru those pts ...............
(x-|z|cos@)(x-1/|z|cos@) + (y-|z|sin@)(y-1/|z|sin@) + k| x y 1 |
| 1/|z| cos@ 1/|z| sin@ 1 |
| |z|cos@ |z|sin@ 1 | <==determinant
for orthogonal intersection
2g1g2 +2f1f2 = c1 + c2
g1, f2, c1 = 0 [for x^2+y^2 =1]
now we have to prove c2 = 0
butt c2 = k(0) +cos^2@ + sin^2@ =1 ...................
so they are orthogonal !!!!!!!!!!!!!!!!!!!!! :((((((((
1now see this solution
the points z0 ,1/z0 and origin are collinear because arg(z0)=arg(1/z0)
now | z0| =r(let)
so 1/|z0|=1/r
now OA.OB= |z0|.1/|z0|=1 = OP2
therefore op is tangent to circle with centre C .
angle OPC=90°
therefore the curves are orthagonal
this is not my solution
66@decoder, at least you should have mentioned that you took the diagram from one of my posts (in goiit probably) :)
341and a problem posted by me and this fantastic solution by jishnudas (kaymant sir's student) :D
1yes sir i have taken this problem from there .............i had mentioned in my post that it is not my solution