Question

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:

• A. $x^2+y^2-4x+8y+12=0$
• B. $x^2+y^2-x+4y-12=0$
• C. $x^2+y^2-\frac{x}{4}+2y-24=0$
• D. $x^2+y^2-4x+9y+18=0$

Answer 1

The correct option is A $x^2+y^2-4x+8y+12=0$

Parametric point of parabola is $(2t^2,4t)$

Centre of given circle is $(0,-6)$

Let the distance be $g\left(t\right)=\sqrt{f\left(t\right)}=\sqrt{4t^4+(4t+6{)}^2}$

To get the minimum value of $g\left(t\right)$, we need to find minimum value of $f\left(t\right)$

$f^{\prime }\left(t\right)=4(4t^3+8t+12)=0$

$\Rightarrow t=-1$

Therefore the required circle centre is $(2,-4)$ and passing through $(0,-6)$

$\Rightarrow x^2+y^2-4x+8y+12=0$