The minimum distance between the curves $x^{2} = 4 y$ and $x^{2} + y^{2} + 18 x + 12 y + 81 = 0$

7 \sqrt{2}

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7 \sqrt{2} - 6

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7 \sqrt{2} + 6

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Solution

The correct option is B $7 \sqrt{2} - 6$
Given, circle is $(x + 9)^{2} + (y + 6)^{2} = 6^{2}$ and parabola is $x^{2} = 4 y$

The minimum distance occurs along the common normal of the two curves.

Equation of the normal will be $x + t y = 2 t + t^{3}$
Since, it's passes through $(- 9, - 6)$
$∴ - 9 - 6 t = 2 t + t^{3} \Rightarrow t^{3} + 8 t + 9 = 0 \Rightarrow (t + 1) (t^{2} - t + 8) = 0 \Rightarrow t = - 1$

So, the nearest point of the parabola is $(2 t, t^{2}) \equiv (- 2, 1)$

So, the distance is $\sqrt{7^{2} + 7^{2}} - 6 = 7 \sqrt{2} - 6$

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The minimum distance between the curves x2=4y and x2+y2+18x+12y+81=0

The minimum distance between the curves $x^{2} = 4 y$ and $x^{2} + y^{2} + 18 x + 12 y + 81 = 0$