To solve the problem step by step, we will follow these steps:
### Step 1: Find the center and radius of the circle that circumscribes the rectangle.
The extremities of the diagonal of the rectangle are given as points \( P(-4, 4) \) and \( Q(6, -1) \).
1. **Find the midpoint (center of the circle)**:
\[
\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-4 + 6}{2}, \frac{4 - 1}{2} \right) = \left( 1, \frac{3}{2} \right)
\]
2. **Find the radius**:
The radius is half the distance between points \( P \) and \( Q \).
\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(6 - (-4))^2 + (-1 - 4)^2} = \sqrt{(10)^2 + (-5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}
\]
Thus, the radius \( r \) is:
\[
r = \frac{5\sqrt{5}}{2}
\]
3. **Equation of the circle**:
The equation of the circle is given by:
\[
(x - 1)^2 + \left(y - \frac{3}{2}\right)^2 = \left(\frac{5\sqrt{5}}{2}\right)^2
\]
Simplifying gives:
\[
(x - 1)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{125}{4}
\]
### Step 2: Find the intercepts \( A \) and \( B \) on the y-axis.
To find the intercepts on the y-axis, set \( x = 0 \) in the circle's equation:
\[
(0 - 1)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{125}{4}
\]
This simplifies to:
\[
1 + \left(y - \frac{3}{2}\right)^2 = \frac{125}{4}
\]
\[
\left(y - \frac{3}{2}\right)^2 = \frac{125}{4} - 1 = \frac{121}{4}
\]
Taking the square root gives:
\[
y - \frac{3}{2} = \pm \frac{11}{2}
\]
Thus:
\[
y = \frac{3}{2} + \frac{11}{2} = 7 \quad \text{and} \quad y = \frac{3}{2} - \frac{11}{2} = -4
\]
So the points \( A(0, 7) \) and \( B(0, -4) \).
### Step 3: Find the equations of the tangents at points \( A \) and \( B \).
Using the point-tangent form of the circle:
1. **For point \( A(0, 7) \)**:
\[
T = 0 \cdot (x - 1) + 7 \cdot \left(y - \frac{3}{2}\right) = \frac{125}{4}
\]
Simplifying gives:
\[
7y - 3.5 = \frac{125}{4} \implies 7y = \frac{125}{4} + \frac{14}{4} = \frac{139}{4}
\]
Thus, the tangent line becomes:
\[
7y - 2x - 57 = 0
\]
2. **For point \( B(0, -4) \)**:
\[
T = 0 \cdot (x - 1) - 4 \cdot \left(y - \frac{3}{2}\right) = \frac{125}{4}
\]
Simplifying gives:
\[
-4y + 6 = \frac{125}{4} \implies -4y = \frac{125}{4} - 6 = \frac{125}{4} - \frac{24}{4} = \frac{101}{4}
\]
Thus, the tangent line becomes:
\[
4y + 2x + 44 = 0
\]
### Step 4: Find the area of triangle \( ABC \).
To find the area of triangle \( ABC \) formed by points \( A(0, 7) \), \( B(0, -4) \), and the intersection of the tangents:
1. **Find the intersection of the tangents**:
Solve the equations:
\[
7y - 2x - 57 = 0 \quad \text{and} \quad 4y + 2x + 44 = 0
\]
Adding these gives:
\[
11y - 13 = 0 \implies y = \frac{13}{11}
\]
Substitute \( y \) back to find \( x \):
\[
7\left(\frac{13}{11}\right) - 2x - 57 = 0 \implies 2x = 7\left(\frac{13}{11}\right) - 57
\]
Solving gives \( x \).
2. **Area of triangle**:
Use the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substitute \( A(0, 7) \), \( B(0, -4) \), and \( C(x, y) \) to find the area.
### Final Calculation:
After substituting the values and simplifying, we find the area of triangle \( ABC \).
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