SATHEE: Parabola 2 Question 7

Parabola 2 Question 7

7. The radius of a circle having minimum area, which touches the curve $y = 4 - x^{2}$ and the lines $y = | x |$ , is

(a) $2 (\sqrt{2} + 1)$

(b) $2 (\sqrt{2} - 1)$

(d) $4 (\sqrt{2} + 1)$

(2017 Main)

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Answer:

Correct Answer: 7. (c)

Solution:

Let the radius of circle with least area be $r$ .

Then, then coordinates of centre $= (0, 4 - r)$ .

Since, circle touches the line $y = x$ in first quadrant

$∴ | \frac{0 - (4 - r)}{\sqrt{2}} | = r \Rightarrow r - 4 = \pm r \sqrt{2}$

$\Rightarrow r = \frac{4}{\sqrt{2} + 1}$ or $\frac{4}{1 - \sqrt{2}}$

But $r \neq \frac{4}{1 - \sqrt{2}}$

$∵ \frac{4}{1 - \sqrt{2}} < 0$

$∴ r = \frac{4}{\sqrt{2} + 1} = 4 (\sqrt{2} - 1)$

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Parabola 2 Question 7

7. The radius of a circle having minimum area, which touches the curve y=4−x2 and the lines y=|x|, is

Answer:

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7. The radius of a circle having minimum area, which touches the curve $y = 4 - x^{2}$ and the lines $y = | x |$ , is