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Radius of the circle that passes through the origin and touches the parabola `y^2=4a x` at the point `(a ,2a)` is `5/(sqrt(2))a` (b) `2sqrt(2)a` `sqrt(5/2)a` (d) `3/(sqrt(2))a`

A

`(5)/(sqrt(2))a`

B

`2sqrt(2)a`

C

`sqrt((5)/(2)a`

D

`(3)/(sqrt(2))a`

Text Solution

Verified by Experts

The correct Answer is:
A
(1)
The equation of tangent to the parabola at (a,2a) is
2ya=2a(x+a)
or y-x-a=0
The passes through (0,0). Therefore,
`a^(2)+4a^(2)+lamda(-a)=0orlamda=5a`
Thus, the required circle is `x^(2)+y^(2)-7ax-ay=0`.
Its radius is
`sqrt((49)/(4)a^(2)+(a^(2))/(4))=(5)/(sqrt(2))a`
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