Centre of the circle inscribed in a rectangle formed by the lines `x^2 - 8x+12 = 0` and `y^2 - 14y +40=0` is

Options

A. (4,7)
B. (7,4)
C. (9,4)
D. (4,9)

Answer

To find the center of the circle inscribed in the rectangle formed by the lines given in the equations \(x^2 - 8x + 12 = 0\) and \(y^2 - 14y + 40 = 0\), we will follow these steps:

### Step 1: Solve the equation for \(x\)
The first equation is:
\[ x^2 - 8x + 12 = 0 \]

We can factor this quadratic equation:
\[ (x - 6)(x - 2) = 0 \]

Thus, the solutions for \(x\) are:
\[ x = 2 \quad \text{and} \quad x = 6 \]

### Step 2: Solve the equation for \(y\)
The second equation is:
\[ y^2 - 14y + 40 = 0 \]

We can factor this quadratic equation:
\[ (y - 10)(y - 4) = 0 \]

Thus, the solutions for \(y\) are:
\[ y = 4 \quad \text{and} \quad y = 10 \]

### Step 3: Determine the vertices of the rectangle
The vertices of the rectangle can be determined using the solutions for \(x\) and \(y\):
1. \( (2, 4) \)
2. \( (2, 10) \)
3. \( (6, 4) \)
4. \( (6, 10) \)

### Step 4: Find the center of the rectangle
The center of the rectangle can be found using the midpoint formula. The coordinates of the center \((x, y)\) can be calculated as:
\[ x = \frac{x_1 + x_2}{2} = \frac{2 + 6}{2} = \frac{8}{2} = 4 \]
\[ y = \frac{y_1 + y_2}{2} = \frac{4 + 10}{2} = \frac{14}{2} = 7 \]

Thus, the center of the rectangle, which is also the center of the inscribed circle, is:
\[ (4, 7) \]

### Final Answer
The center of the circle inscribed in the rectangle is:
\[ \boxed{(4, 7)} \]

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