The radical centre of the circles drawn on the focal chords of `y^(2)=4ax` as diameters, is

Options

A. (-a, 0)
B. (a, 0)
C. (0, 0)
D. (a, a)

Answer

C

Explanation

Let `P(at^(2), 2at)" and Q"(a/t^(2),(-2a)/t)` be the equation of the circle with PQ as a diameter is
`(x-at^(2))(x-a/t^(2))+(y-2at)(y+(2a)/t)=0`
`rArr" "x^(2)+y^(2)-ax(t_(1)^(2)+1/t^(2))-2ay(t+1/t)-3a^(2)=0" ...(i)"`
Similarly, the equation of the circle discribed on the focal choed RS as diameter is
`x^(2)+y^(2)-ax(t_(1)^(2)+1/t_(1)^(2))-2ay(t_(1)+1/t_(1))-3a^(2)=0" ...(ii)"`
Where `R(at_(1)^(2),2at_(1))" and S"(a/t_(1)^(2),-(2a)/t_(1))` are the end-point of another focal chord.
The radical axis of circles (i) and (ii) is of the form
`lamdax+muy=0` which is a line passing through the origin. Thus, if circles are drawn on the focal chords of `y^(2)=4ax` as diameters, them their radical axes taken in pairs pass through the origin.
Hence, theradical centre is at (0, 0).