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A square is incribed in a circle x^2+y^2...

A square is incribed in a circle `x^2+y^2-6x+8y-103=0` such that its sides are parallel to co-ordinate axis then the distance of the nearest vertex to origin, is equal to (A) `13` (B) `sqrt127` (C) `sqrt41` (D) `1`

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To solve the problem, we need to find the distance of the nearest vertex of a square inscribed in a given circle to the origin. The circle is defined by the equation: \[ x^2 + y^2 - 6x + 8y - 103 = 0 \] ### Step 1: Rewrite the Circle Equation in Standard Form We start by rewriting the equation of the circle in standard form. We will complete the square for both \(x\) and \(y\). ...
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