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  •  Let P be the point on the parabola,  , which is at a minimum distance from the cent

 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}+(y+6^{2})=1 .  Then, the equation of the circle, passing through C and having its center at P is

Option: 1

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


Option: 2

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


Option: 3

x^{2}+y^{2}-\frac{x}{4}+2y-24=0


Option: 4

x^{2}+y^{2}-4x+9y+18=0


Answers (1)

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

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

Now, let

Let   \begin{aligned} & D=C P \\ \Rightarrow & D=\sqrt{\left(2 t^2\right)^2+(4 t+6)^2} \\ \Rightarrow & D=\sqrt{4 t^4+16 t^2+36+48 t} \\ \Rightarrow & D^2=4 t^4+16 t^2+36+48 t \\ \Rightarrow & F(t)=4 t^4+16 t^2+36+48 t \end{aligned}

For minimum,

\begin{aligned} & F^{\prime}(t)=0 \\ \Rightarrow & 16 t^3+32 t+48=0 \\ \Rightarrow & t^3+2 t+3=0 \\ \Rightarrow & (t+1)\left(t^2-t+3\right)=0 \\ \Rightarrow & t=-1 \end{aligned}

Thus, coordinate of point P are (2,-4)

C P=\sqrt{2^2+(-4+6)^2}=\sqrt{4+4}=1 \sqrt{2}

Hence, the required equation of circle is

\begin{aligned} & (x-2)^2+(y+4)^2=(2 \sqrt{2})^2 \\ & \Rightarrow x^2+y^2-4 x+8 y+12=0 \end{aligned}

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Kshitij

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