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

Options

A. `x^(2)+y^(2)-x+4y-12=0`
B. `x^(2)+y^(2)=(1)/(4)x+2y-24=0`
C. `x^(2)+y^(2)-4x+9y+12=0`
D. `x^(2)+y^(2)-4x+8y+12=0`

Answer

D

Explanation

Let `P(2t^(2), 4t)` be a point on `y^(2)=8x`. The coordinates of the centre C are(0, -6).
`:." "CP^(2)=4t^(4)+(4t+6)^(2)`
`"Let z =CP"^(2)` Then,
`z=4t^(4)+4(2t+3)^(2)`
`:." "(dz)/(dt)=16t^(3)+16(2t+3)" and "(d^(2)z)/(dt^(2))=48t^(2)+32`
For maximum or minimum values of z, we must have
`(dz)/(dt)=0rArr16(t^(3)+2t+3)=0`
`rArr" "(t+1)(t^(2)-t+3)=0rArrt=-1`
Clearly, `((d^(2)z)/(dt^(2)))_(t=-1)=80gt." So, z or CP is minimum when"`
t=-1. So, the coordinates of P are (2, -4).
The equation of the circle having centre at P(2, -4) adn passing through C(0, -6) is
`(x-2)^(2)+(y+4)^(2)=(0-2)^(2)+(-6+4)^(2)`
`"or, "x^(2)+y^(2)-4x+8y+12=0`
ALITER Substituting t=2 in (i), the equation of the normal at P(4, 2) is y + 2x=12 whose x intercep is 6.
The tangent to the circle at Q is perpendicular to the normal `y+2x==12`, So, the slope of the tangent is `1/2`.