Let P be the point on parabola `y^2=4x` which is at the shortest distance from the center S of the circle `x^2+y^2-4x-16y+64=0` let Q be the point on the circle dividing the line segment SP internally. Then

Answer

`A to p; B to q: C to s; D to r`

Explanation

since, equation of normal to the parabola `y^(2) = 4ax` is `y + xt = 2at + at^(3)` passes through (3,0)
`implies 3t = 2t + t^(2)`
`implies t = 0, 1 -1`
`:. ` Coordinates of the normals are `P(1,2), Q(0,0) R(1,-2)`
Thus,
A. Area of `Delta PQR = (1)/(2) xx 1 xx 4 = 2`
C. Centriod of `Delta PQR = ((2)/(3), 0)`
Equation of circle passing through P,Q,R is
(x - 1)(x -1) + (y - 2) (y + 2) + `lambda` (x - 1) = 0
`implies 1 - 4 - lambda = 0`
`implies lambda = - 3`
`:.` Required equation of circle is
`x^(2) + y^(2) - 5x = 0`
`:.` Centre `((5)/(2) , 0)` and radius `(5)/(2)`.