Discussion Forum

Let P be the point on the parabola, y²=8x, which is 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

Answer:

Explanation:

Centre of circle $x^{2}+(y+6)^{2}=1 $ is C (0,-6)

 Let the coordinates of point $(2t^{2},4t)$

Now let D=CP

   $=\sqrt{(2t^{2})^{2}-(4t+6)^{2}}$

$\Rightarrow D=\sqrt{4t^{4}-16t^{2}+36+48t}$

squaring on both sides

$\Rightarrow D^{2}(t)=4t^{4}+16t^{2}+48t+36$

 Let $F(t)=4t^{4}+16t^{2}+48t+36$

 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 

$(x-2)^{2}+(y+4)^{2}=(2\sqrt{2}^{2})$

$x^{2}+4-4x+y^{2}+16+8y=8$

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