One diagonal of a square is the portion of x-axis intercepted by the circle `x^2+y^2-4x+6y-12=0` The extremity of the other diagonal is

Options

A. (2,4)
B. (2,-4)
C. (-2,-4)
D. none of these

Answer

To solve the problem, we need to find the extremities of the other diagonal of a square, given that one diagonal lies along the x-axis and is defined by the points where the circle intersects the x-axis.

### Step-by-Step Solution:

1. **Identify the Circle Equation**:
The given equation of the circle is:
\[ x^2 + y^2 - 4x - 6y - 12 = 0 \]

2. **Find the x-intercepts**:
To find the x-intercepts, set \( y = 0 \) in the circle's equation:
\[ x^2 - 4x - 12 = 0 \]

3. **Solve the Quadratic Equation**:
Factor the quadratic equation:
\[ (x - 6)(x + 2) = 0 \]
This gives us the solutions:
\[ x = 6 \quad \text{and} \quad x = -2 \]
Therefore, the x-intercepts are the points \( (6, 0) \) and \( (-2, 0) \).

4. **Find the Midpoint of the Diagonal**:
The midpoint \( O \) of the diagonal \( AC \) (where \( A = (6, 0) \) and \( C = (-2, 0) \)) can be calculated as:
\[ O = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{6 + (-2)}{2}, \frac{0 + 0}{2} \right) = \left( \frac{4}{2}, 0 \right) = (2, 0) \]

5. **Calculate the Length of the Diagonal**:
The length of diagonal \( AC \) is the distance between the points \( A \) and \( C \):
\[ AC = |6 - (-2)| = 6 + 2 = 8 \]
Since the diagonals of a square are equal, \( BD = AC = 8 \).

6. **Find the Length of Half the Diagonal**:
The length of half the diagonal \( OB \) is:
\[ OB = \frac{AC}{2} = \frac{8}{2} = 4 \]

7. **Determine the Extremities of the Other Diagonal**:
Since \( O = (2, 0) \) and the length \( OB = 4 \), we can find the points \( B \) and \( D \) by moving vertically (since the other diagonal is perpendicular to the x-axis).
The coordinates of points \( B \) and \( D \) will be:
\[ B = (2, 4) \quad \text{and} \quad D = (2, -4) \]

### Final Answer:
The extremities of the other diagonal are \( (2, 4) \) and \( (2, -4) \).

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