Let P be the point on the parabola, `y^2=8x` which is at a minimum distance from the centre C of the circle,`x^2+(y+6)^2=1.` Then the equation of the circle, passing through C and having its centre at P is :

Options

A. `x^2+y^2-4x+8y+12=0`
B. `x^2+y^2-x+4y-12=0`
C. `x^2+y^2-x/4+2y-24=0`
D. `x^2+y^2-4x+9y+18=0`

Answer

A

Explanation

Centre of circle `x^(2) + (y + 6)^(2) = 1` is `C(0, -6)`
Let the coordinates of point P be `(2t^(2), 4t)`
Now, let D = CP
`= sqrt((2t^(2))^(2) + (4 t + 6)^(2))`
`implies D = sqrt(4 t^(4) + 16 t^(2) + 36 + 48 t)`
Squaring on both sides
`implies D^(2) (t) = 4 t^(4) + 16 t^(2) + 48 t + 36`
Let `F (t) = 4 t^(4) + 16 t^(2) + 48 t + 36`
For minimum, `F' (t) = 0`
`implies 16 t^(3) + 32 t + 48 = 0`
`implies t^(3) + 2t + 3 = 0`
`implies (t + 1) (t^(2) - t + 3) = 0 implies t = - 1`
Thus, coordinate of point P are (2, -4)
Now, `CP = sqrt(2^(2) + (-4 + 6)^(2)) = sqrt(4 + 4) = 2 sqrt(2)`
Hence, the required equation of circle is
`(x - 2)^(2) + (y + 4)^(2) = (2 sqrt(2))^(2)`
`implies x^(2) + 4 - 4x + y^(2) + 1 + 8y = 8`
`implies x^(2) + y^(2) - 4x + 8y + 12 = 0`