JEE Main 2017Session, Shift 1MathematicsMediumCoordinate Geometry478× since 2002The radius of a circle, having minimum area, which touches the curve y = 4 – x² and the lines, y = |x| is :AOption A. 2(2−1)2\left( {\sqrt 2 - 1} \right)2(2−1)BOption B. 4(2−1)4\left( {\sqrt 2 - 1} \right)4(2−1)COption C. 4(2+1)4\left( {\sqrt 2 + 1} \right)4(2+1)DOption D. 2(2+1)2\left( {\sqrt 2 + 1} \right)2(2+1)CheckSkip · show solution
