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`

Explanation

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 steps to solve this are as follows:

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 - 6x + 8y - 103 = 0 \]

We can rewrite this in standard form by completing the square.

...