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`

Options

A. 13
B. `sqrt(137)`
C. `6`
D. `sqrt(41)`

Answer

To solve the problem, we need to find the distance of the nearest vertex of a square inscribed in a circle to the origin. The circle is defined by the equation:

\[ x^2 + y^2 - 6x + 8y - 103 = 0 \]

### Step 1: Rewrite the equation of the circle in standard form

To find the center and radius of the circle, we will complete the square for the \(x\) and \(y\) terms.

1. Rearranging the equation:
\[ x^2 - 6x + y^2 + 8y = 103 \]

2. Completing the square for \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

3. Completing the square for \(y\):
\[ y^2 + 8y = (y + 4)^2 - 16 \]

4. Substitute back into the equation:
\[ (x - 3)^2 - 9 + (y + 4)^2 - 16 = 103 \]
\[ (x - 3)^2 + (y + 4)^2 = 128 \]

### Step 2: Identify the center and radius of the circle

From the equation \((x - 3)^2 + (y + 4)^2 = 128\), we can identify:

- Center: \( (3, -4) \)
- Radius: \( r = \sqrt{128} = 8\sqrt{2} \)

### Step 3: Determine the side length of the inscribed square

Since the square is inscribed in the circle and its sides are parallel to the coordinate axes, the diagonal of the square is equal to the diameter of the circle.

1. Diameter of the circle:
\[ D = 2r = 2 \times 8\sqrt{2} = 16\sqrt{2} \]

2. Let the side length of the square be \(s\). The relationship between the side length and the diagonal is given by:
\[ s\sqrt{2} = D \]
\[ s\sqrt{2} = 16\sqrt{2} \]
\[ s = 16 \]

### Step 4: Find the vertices of the square

The vertices of the square can be calculated using the center and the side length. The center is at \( (3, -4) \), and since the sides are parallel to the axes, the vertices will be:

1. Top right vertex:
\[ \left(3 + \frac{s}{2}, -4 + \frac{s}{2}\right) = \left(3 + 8, -4 + 8\right) = (11, 4) \]

2. Top left vertex:
\[ \left(3 - \frac{s}{2}, -4 + \frac{s}{2}\right) = \left(3 - 8, -4 + 8\right) = (-5, 4) \]

3. Bottom left vertex:
\[ \left(3 - \frac{s}{2}, -4 - \frac{s}{2}\right) = \left(3 - 8, -4 - 8\right) = (-5, -12) \]

4. Bottom right vertex:
\[ \left(3 + \frac{s}{2}, -4 - \frac{s}{2}\right) = \left(3 + 8, -4 - 8\right) = (11, -12) \]

### Step 5: Calculate the distances from the origin to each vertex

Using the distance formula \(d = \sqrt{x^2 + y^2}\):

1. Distance to \( (11, 4) \):
\[ d_A = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137} \]

2. Distance to \( (-5, 4) \):
\[ d_B = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \]

3. Distance to \( (-5, -12) \):
\[ d_C = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

4. Distance to \( (11, -12) \):
\[ d_D = \sqrt{11^2 + (-12)^2} = \sqrt{121 + 144} = \sqrt{265} \]

### Step 6: Identify the minimum distance

The distances calculated are:
- \(d_A = \sqrt{137}\)
- \(d_B = \sqrt{41}\)
- \(d_C = 13\)
- \(d_D = \sqrt{265}\)

The minimum distance is \(d_B = \sqrt{41}\).

### Final Answer

The distance of the nearest vertex to the origin is:

\[ \sqrt{41} \]