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The radius of a circle, having minimum area, which touches the curve y=4−x2

and the lines, y=\left | x \right | is:

  • Option 1)

    2\left ( \sqrt{2}-1 \right )

  • Option 2)

    4\left ( \sqrt{2}-1 \right )

  • Option 3)

    4\left ( \sqrt{2}+1 \right )

  • Option 4)

    2\left ( \sqrt{2}+1 \right )

 

Answers (1)

best_answer

As we learnt in

Standard equation of parabola -

x^{2}=-4ay

- wherein

 

 and concept

Perpendicular distance of a point from a line -

\rho =\frac{\left | ax_{1}+by_{1}+c\right |}{\sqrt{a^{2}+b^{2}}}

 

 

- wherein

\rho  is the distance from the line ax+by+c=0 .

 This figure shows the circle with least area satisfying the conditions.

Circle touches y=x;

\therefore \left | \frac{0-(4-2)}{\sqrt{2}+1} \right |=r (distance\ of\ point\ from\ a\ line)

r=\frac{4}{\sqrt{2}+1}

\Rightarrow r={4}({\sqrt{2}-1})


Option 1)

2\left ( \sqrt{2}-1 \right )

Incorrect

Option 2)

4\left ( \sqrt{2}-1 \right )

Correct

Option 3)

4\left ( \sqrt{2}+1 \right )

Incorrect

Option 4)

2\left ( \sqrt{2}+1 \right )

Incorrect

Posted by

Aadil

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