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.

Options

A. (9,6)
B. `(2,sqrt(8))`
C. (4,4)
D. `(3,2sqrt(3))`

Answer

A, C

Explanation

1,3
Let the possible be `(t^(2),2t)`. The equation of tangent at this point is `yt=x+t^(2)`
It must pass through (6,5), since the normal to the circle always passes through its center. Therefore,
`t^(2)-5t+6=0`
or t=2,3
So, the possible point are (4,4) and (9,6).