Question:
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 :
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 :
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Find the centre and radius; the vertices are the centre shifted by (plus or minus half-side) in each coordinate.
Updated On: Nov 5, 2026
- \(\sqrt{137}\)
- \(6\)
- \(\sqrt{41}\)
- \(13\)
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The Correct Option is C
Solution and Explanation
Circle: \((x-3)^2 + (y+4)^2 = 9 + 16 + 103 = 128\). Centre \((3,-4)\), radius \(8\sqrt2\).
A square with sides parallel to the axes, inscribed in the circle, has diagonal \(= 2R = 16\sqrt2\), so side \(=16\) and half-side \(=8\).
Vertices: \((3\pm8,\,-4\pm8)\) = \((11,4),\ (11,-12),\ (-5,4),\ (-5,-12)\).
Nearest to the origin is \((-5,4)\): distance \(=\sqrt{25+16}=\sqrt{41}\).
Answer: (3) \(\sqrt{41}\)
A square with sides parallel to the axes, inscribed in the circle, has diagonal \(= 2R = 16\sqrt2\), so side \(=16\) and half-side \(=8\).
Vertices: \((3\pm8,\,-4\pm8)\) = \((11,4),\ (11,-12),\ (-5,4),\ (-5,-12)\).
Nearest to the origin is \((-5,4)\): distance \(=\sqrt{25+16}=\sqrt{41}\).
Answer: (3) \(\sqrt{41}\)
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