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

JEE · Math · previous-year question

  1. A.$$2\left( {\sqrt 2 - 1} \right)$$
  2. B.$$4\left( {\sqrt 2 - 1} \right)$$correct
  3. C.$$4\left( {\sqrt 2 + 1} \right)$$
  4. D.$$2\left( {\sqrt 2 + 1} \right)$$

Answer

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

Explanation

Let the radius of circle with least area be r. Then, the coordinate of the center = (0, b) $$ \therefore $$ The equation of circle be x2 + (y – b)2 = r2 Distance of perpendiculur from (0, 4) to y = x line = r $$ \Rightarrow $$ $$\left| {{{ - b} \over {\sqrt 2 }}} \right| = r$$ $$ \Rightarrow $$ b = $${\sqrt 2 r}$$ Circle passes through (0, 4), $$ \therefore $$ 0 + (4 – b)2 = r2 $$ \Rightarrow $$ 4 - b = r $$ \Rightarrow $$ 4 - $${\sqrt 2 r}$$ = r $$ \Rightarrow $$ r = $${4 \over {\sqrt 2 + 1}}$$ = $$4\left( {\sqrt 2 - 1} \right)$$

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