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Let P be the point on the parabola, y...

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 : (1) `x^2+y^2-4x+8y+12=0` (2) `x^2+y^2-x+4y-12=0` (3) `x^2+y^2-x/4+2y-24=0` (4) `x^2+y^2-4x+9y+18=0`

A

`x^(2)+y^(2)-x+4y-12=0`

B

`x^(2)+y^(2)-(x)/(4)+2y-24=0`

C

`x^(3)+y^(2)-4x+9y-18=0`

D

`x^(2)+y^(2)-4x+8y-12=0`

Text Solution

Verified by Experts

The correct Answer is:
D
4 Point P lies on the common normal to parabola at point P and circle.

For `y^(2)=8x`, point P is `(2t^(2),4t)`
Equation of normal at this P is `y=-tx+4t+2t^(3)`.
This is also normal to the given circle, so it must pass through the center of the circle.
`:." "-6=4t+2t^(3)`
`:." "t^(3)+2t+3=0`
`:." "(t+1)(t^(2)-t+3)=0`
`:." "t=-1`
`:." Point P is "(2,-4)`
`:." Equation of required circle is "(x-2)^(2)+(y+4)^(2)=r^(2)=8`
`"Or "x^(2)+y^(2)-4x+8y+12=0`
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