Circle and a parabola

2√13

one thing, the chord of contact will be symmetrical wrt x-axis..
(since both circle n parabola are symmetrical to x-axis..)

let radius be 'r'.

take the two extremities of chord of contact as A(rcosθ1, rsinθ1)
and B(rcosθ2, rsinθ2).

but since the line is symm with x-axis, θ2=θ1.
so the two points become A(rcosθ1, rsinθ1) and B(rcosθ1, -rsinθ1).

given that the ch of contact subtends a r8 angle at vertex i.e. (0,0)

so m1.m2 =-1.
from here we get sin²Î¸1/cos²Î¸1 = 1
=> tanθ= +- 1. => θ=Π/4, -Π/4.

These points also lie on the parabola.

so (rcos pi/4 , rsin pi/4) must satisfy parabola eqn.

then r= 4√2.

you should note that the centre of the circle is not origin but is at (-2,0)

:(
yeah i missed that...

we have to take the points -2+rcosθ1,0+rsinθ1 .

shit.

Let the eqn of common chord be x=k .........(why?)

homogenize it with y²=4x

to get y²-4x²/k

Now 1-4/k=0...
so k=4

so chord is x=4...
It cuts parabola at (4,4)

radius of circle is distance b/w (-2,0) and (4,4) which is 2√13....

wow.. gr8 method :)

mughe yeh samaj me nahi araha ki aap logon ne kaise centre of the circle ko (-2, 0) liya ?

the one shown in the diagram is also possbile no?

mera doubt clear karo koi ? sorry if it's dumb.

here's the figure.
this is also possible no?

Hey I think the question was edited after sonls.... so the solns are not looking appropriate