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The parabola y^2=4x and the circle havin...

The parabola `y^2=4x` and the circle having its center at 6, 5) intersect at right angle. Then find the possible points of intersection of these curves.

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Since circle intersects parabola orthogonally, the tangent to parabola at its point of intersection with circle must be normal to the circle.
Normal to the circle passes through its centre.
Let the point of intersection of circle and parabola be `P(t^(2),2t)`.
Equation of tangent at P to the parabola is `yt=x+t^(2)`.
This is normal to circle.
So, it must pass through centre of the circle (6,5).
`:." "t^(2)-5t+6=0`
`:." "t=2,3`
So, possible point on the parabola are (4,4) and (9,6).
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