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 ${}^{\prime }{\left(0,-6\right)}^{\prime }$ , 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 $\left(0,-6\right)$
$\Rightarrow m^3+2m-3=0\Rightarrow m=1$
$P\left(a{m}^2,-2am\right)=P(2,-4)$
Equation of circle with centre $P$ and passes through ${}^{\prime }{C}^{\prime }$ is
${\left(x-2\right)}^2+{\left(y+4\right)}^2=8$
$\Rightarrow x^2+y^2-4x+8y+12=0$