Statement I The line 3x-4y=7 is a diameter of the circle `x^(2)+y^(2)-2x+2y-47=0`
Statement II Normal of a circle always pass through centre of circle

Options

A. Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I
B. Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I
C. Statement I is true, Statement II is false
D. Statement I is false, Statement II is true

Answer

To determine the validity of the statements regarding the circle and the line, we will analyze each statement step by step.

### Step 1: Identify the center of the circle
The given equation of the circle is:
\[ x^2 + y^2 - 2x + 2y - 47 = 0 \]

To find the center, we can rewrite the equation in standard form by completing the square.

1. Rearranging the equation:
\[ (x^2 - 2x) + (y^2 + 2y) = 47 \]

2. Completing the square:
- For \(x^2 - 2x\), we take half of -2 (which is -1), square it (getting 1), and add it to both sides:
\[ (x - 1)^2 - 1 \]
- For \(y^2 + 2y\), we take half of 2 (which is 1), square it (getting 1), and add it to both sides:
\[ (y + 1)^2 - 1 \]

3. Now substituting back:
\[ (x - 1)^2 - 1 + (y + 1)^2 - 1 = 47 \]
\[ (x - 1)^2 + (y + 1)^2 = 49 \]

From this, we can see that the center \(C\) of the circle is:
\[ C(1, -1) \]