Parabola 2 Question 28

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

(c) $4(\sqrt{2}-1)$

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

(2017 Main)

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Answer:

Correct Answer: 7. (c)

Solution:

  1. 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

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

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

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

$$ \because \frac{4}{1-\sqrt{2}}<0 $$

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



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