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

Answer: (A)

Normal at $P(a{t}^2,2at)$ is $y+tx=2at+a{t}^3$
Given it passes $(0,-6)$
$\Rightarrow -6=2at+a{t}^3(a=2)$
$-6=4t+2t^3$
$t^3+2t+3=0$
$t=-1$
so, $P(a,-2a)=(2,-4).[a=1)$
radius of circle $=CP=\sqrt{2^2+(-4+6{)}^2}=2\sqrt{2}$
Circle is $(x-2{)}^2+(y+4{)}^2=(2\sqrt{2}{)}^2$
$x^2+y^2-4x+8y+12=0$