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 is the portion of the x-axis intercepted by the circle defined by the equation \(x^2 + y^2 - 4x + 6y - 12 = 0\).

### Step-by-Step Solution:

**Step 1: Rewrite the Circle Equation**
We start by rewriting the circle equation in standard form. The given equation is:
\[ x^2 + y^2 - 4x + 6y - 12 = 0 \]
We can complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 4x \rightarrow (x - 2)^2 - 4 \]

For \(y\):
\[ y^2 + 6y \rightarrow (y + 3)^2 - 9 \]

Substituting these back into the equation gives:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 25 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 = 25 \]
This represents a circle with center \((2, -3)\) and radius \(5\).

**Step 2: Find the Points of Intersection with the x-axis**
To find the points where the circle intersects the x-axis, we set \(y = 0\):
\[ (x - 2)^2 + (0 + 3)^2 = 25 \]
\[ (x - 2)^2 + 9 = 25 \]
\[ (x - 2)^2 = 16 \]
Taking the square root gives:
\[ x - 2 = 4 \quad \text{or} \quad x - 2 = -4 \]
Thus,
\[ x = 6 \quad \text{or} \quad x = -2 \]
So, the points of intersection are \((-2, 0)\) and \((6, 0)\).

**Step 3: Identify the Diagonal of the Square**
The points \((-2, 0)\) and \((6, 0)\) are the endpoints of one diagonal of the square. Let’s denote these points as \(A(-2, 0)\) and \(C(6, 0)\).

**Step 4: Find the Midpoint of the Diagonal**
The midpoint \(O\) of diagonal \(AC\) can be calculated as:
\[ O = \left( \frac{-2 + 6}{2}, \frac{0 + 0}{2} \right) = \left( 2, 0 \right) \]

**Step 5: Determine the Length of the Diagonal**
The length of diagonal \(AC\) is:
\[ AC = \sqrt{(6 - (-2))^2 + (0 - 0)^2} = \sqrt{8^2} = 8 \]
Since the diagonals of a square are equal, the length of diagonal \(BD\) is also \(8\).

**Step 6: Find the Coordinates of Points B and D**
Since \(BD\) is perpendicular to \(AC\) and bisects it at point \(O(2, 0)\), we can find the coordinates of points \(B\) and \(D\) by moving vertically from point \(O\) by half the length of the diagonal:
\[ \text{Half of diagonal} = \frac{8}{2} = 4 \]
Thus, 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:
\[ B(2, 4) \quad \text{and} \quad D(2, -4) \]