devarshi-dt-logo

Question:

Let P be the point on the parabola, y²=8x which is at a minimum distance from the centre C of the circle, x²+(y+6)²=1. Then the equation of the circle, passing through C and having its centre at P is:

x²+y²−x+4y−12=0

x²+y²−x⁴+2y−24=0

x²+y²−8x+8y+12=0

x²+y²−8x+9y+18=0

Solution:

Parametric point of parabola is (2t²,4t)
Centre of given circle is (0,-6)
Let the distance be g(t)=√f(t)=√(2t²)²+(4t+6)²
To get the minimum value of g(t), we need to find minimum value of f(t)
f'(t)=4(4t³+8t+12)=0 ⇒t=-1
Therefore the required circle centre is (2,-4) and passing through (0,-6)
⇒x²+y²+8x+8y+12=0