Question

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

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

Answer 1

Answer: (C)

$y^{2}=8x$ is the equation of the given parabola. If $P$ is a point at a minimum distance from $'\left(0 , - 6\right)'$ , then it should be normal to the parabola at $P$ .

Normal to parabola $y^{2}=8x$ is
$y=mx-2\cdot 2\cdot m-2\cdot m^{3}$
It passes through $\left(0 , - 6\right)$
$\Rightarrow \, m^{3}+2m-3=0\Rightarrow m=1 \, $
$P\left(a m^{2} , - 2 a m\right)=P\left(\right.2,-4\left.\right)$
Equation of circle with centre $P$ and passes through $'C'$ is
$\left(x - 2\right)^{2}+\left(y + 4\right)^{2}=8$
$\Rightarrow x^{2}+y^{2}-4x+8y+12=0$