JEE PYQ: Circle Question 38
Question 38 - 2019 (11 Jan Shift 1)
A square is inscribed in the circle $x^2 + y^2 - 6x + 8y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is:
(1) $6$
(2) $\sqrt{137}$
(3) $\sqrt{41}$
(4) $13$
Show Answer
Answer: (3)
Solution
Center $(3, -4)$, $r = \sqrt{9+16+103} = 8\sqrt{2}$. Side of square $= \sqrt{2} \times r = 16$… Actually $r = \sqrt{128} = 8\sqrt{2}$. Side $= r\sqrt{2} = 16$. Vertices: $(3 \pm 8, -4 \pm 8)$. Nearest to origin: $(-5, 4)$, distance $= \sqrt{25+16} = \sqrt{41}$.
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