Q

# Help me answer: - Co-ordinate geometry - JEE Main

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

and the lines, is:

• Option 1)

• Option 2)

• Option 3)

• Option 4)

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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)

Incorrect

Option 2)

Correct

Option 3)

Incorrect

Option 4)

Incorrect

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