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

Centre of the circle $x^2+(y+6{)}^2=1$ is $C(0,-6)$
Let the coordinates of point P be $(2t^2,4t)$
Now, let
$D=CP=\sqrt{(2t^2{)}^2+(4t+6{)}^2}\Rightarrow D=\sqrt{4t^4+16t^2+36+48t}$
Squaring on both sides
$\Rightarrow D^2\left(t\right)=4t^4+16t^2+48t+36$
Let $F\left(t\right)=4t^4+16t^2+48t+36$
For minimum F'(t)=0
$\Rightarrow 16t^3+32t+48=0\Rightarrow t^3+2t+3=0\Rightarrow (t+1)(t^2-t+3)=0\Rightarrow t=-1$
Thus, coordinate of point P are $(2,-4)$
Now, $CP=\sqrt{2^2+(-4+6{)}^2}=\sqrt{4+4}$
$=2\sqrt{2}$
Hence, the required equation of circle is
$(x-2{)}^2+(y+4{)}^2=(2\sqrt{2}{)}^2\Rightarrow x^2+4-4x+y^2+16+8y=8\Rightarrow x^2+y^2-4x+8y+12=0$