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Step 1: Rewrite the circle in standard form
To find the center and radius of the circle
● Given equation:
$ x^2 + y^2 - 6x + 8y - 103 = 0 $
● Complete the square for both x and y:
$ x^2 - 6x + 9 + y^2 + 8y + 16 = 103 + 9 + 16 $
$ (x - 3)^2 + (y + 4)^2 = 128 $
Thus, the center is $(3,\, -4)$ and the radius $R$ is $ \sqrt{128} = 8\sqrt{2} $.
Step 2: Determine the side length of the inscribed square
Since the square is axis-aligned and inscribed in the circle, its center coincides with the circle’s center. Each vertex of the square is at the same distance (equal to the radius $8\sqrt{2}$) from the center.
● If the side of the square is $s$, the distance from the center to any vertex is half the diagonal, i.e., $ \frac{s}{\sqrt{2}} $.
● Set this distance equal to the circle’s radius:
$ \frac{s}{\sqrt{2}} = 8\sqrt{2} \implies s = 8\sqrt{2} \cdot \sqrt{2} = 16. $
Step 3: Find the coordinates of the vertices
The center of the square (and circle) is $(3,\, -4)$. If the side is 16, then half the side is 8. The vertices of the square are
● $(3 + 8,\,-4 + 8) = (11,\,4),$
● $(3 + 8,\,-4 - 8) = (11,\,-12),$
● $(3 - 8,\,-4 + 8) = (-5,\,4),$
● $(3 - 8,\,-4 - 8) = (-5,\,-12).$
Step 4: Calculate the distances of these vertices from the origin
● Distance to $(11,\,4)$: $ \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137}. $
● Distance to $(11,\,-12)$: $ \sqrt{11^2 + (-12)^2} = \sqrt{121 + 144} = \sqrt{265}. $
● Distance to $(-5,\,4)$: $ \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}. $
● Distance to $(-5,\,-12)$: $ \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13. $
The nearest vertex to the origin is $(-5,\,4)$ with distance $ \sqrt{41} $.
Hence, the required distance of the nearest vertex of the square from the origin is $ \sqrt{41}. $
