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QJEEM19Ma773a: JEE Main 2019 Mathematics - Class 11 JEEM19Ma773a Solution: A square is inscribed in the circle x 2 + y 2 – 6x + 8y – 103 = …

A square is inscribed in the circle x2 + y2 – 6x + 8y – 103 = 0 with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is :

Multiple Choice Options

  • A 137\sqrt {137}
  • B 6
  • C 41\sqrt {41}
  • D 13

Step-by-Step Verified Solution

JEE Main 2019 (Online) 11th January Morning Slot Mathematics - Circle Question 122 English Explanation
R =9+16+103=82= \sqrt {9 + 16 + 103} = 8\sqrt 2

OA =13= 13

OB =265= \sqrt {265}

OC =137= \sqrt {137}

OD =41= \sqrt {41}

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