Find the equation of the circle that circumscribes the triangle determined by the lines y = 1 - 0, x = y - 0, and x + y + 2 = 0.

To find the equation of the circle that circumscribes the triangle formed by the lines \( y = 1 \), \( x = y \), and \( x + y + 2 = 0 \), we can follow these steps:

### 1. **Find the points of intersection:**

   - **Intersection of \( y = 1 \) and \( x = y \):**
     \[
     y = 1 \quad \text{and} \quad x = y \implies x = 1
     \]
     So, the point is \( (1, 1) \).

   - **Intersection of \( y = 1 \) and \( x + y + 2 = 0 \):**
     \[
     y = 1 \quad \text{and} \quad x + 1 + 2 = 0 \implies x = -3
     \]
     So, the point is \( (-3, 1) \).

   - **Intersection of \( x = y \) and \( x + y + 2 = 0 \):**
     \[
     x = y \quad \text{and} \quad x + x + 2 = 0 \implies 2x = -2 \implies x = -1 \implies y = -1
     \]
     So, the point is \( (-1, -1) \).

### 2. **Use the circumcenter formula:**

The circumcenter of a triangle is the center of the circumscribed circle and can be found by using the formula for the circumcenter in terms of the vertices of the triangle.

Let's denote the points as \( A(1, 1) \), \( B(-3, 1) \), and \( C(-1, -1) \).

### 3. **Find the perpendicular bisectors:**

To find the circumcenter, we can find the intersection of the perpendicular bisectors of the sides of the triangle.

- The midpoint of \( AB \) is:
  \[
  \left(\frac{1 - 3}{2}, \frac{1 + 1}{2}\right) = \left(-1, 1\right)
  \]
  The slope of \( AB \) is:
  \[
  m_{AB} = \frac{1 - 1}{1 + 3} = 0
  \]
  The perpendicular bisector of \( AB \) is vertical, passing through \( (-1, 1) \).

- The midpoint of \( BC \) is:
  \[
  \left(\frac{-3 - 1}{2}, \frac{1 - 1}{2}\right) = \left(-2, 0\right)
  \]
  The slope of \( BC \) is:
  \[
  m_{BC} = \frac{1 - (-1)}{-3 + 1} = -1
  \]
  The perpendicular slope is \( 1 \), and the perpendicular bisector is:
  \[
  y - 0 = 1(x + 2) \implies y = x + 2
  \]

### 4. **Find the intersection of perpendicular bisectors:**

Since \( x = -1 \) from the first perpendicular bisector, substituting it into the second:
\[
y = -1 + 2 = 1
\]
So, the circumcenter (center of the circle) is at \( (-1, 1) \).

### 5. **Find the radius:**

The radius \( R \) is the distance from the circumcenter \( (-1, 1) \) to any of the vertices, say \( (1, 1) \):
\[
R = \sqrt{(1 + 1)^2 + (1 - 1)^2} = \sqrt{4 + 0} = 2
\]

### 6. **Equation of the circle:**

The equation of a circle with center \( (h, k) \) and radius \( R \) is:
\[
(x - h)^2 + (y - k)^2 = R^2
\]
Substituting \( h = -1 \), \( k = 1 \), and \( R = 2 \):
\[
(x + 1)^2 + (y - 1)^2 = 4
\]

### Final Answer:

The equation of the circumscribed circle is:
\[
\boxed{(x + 1)^2 + (y - 1)^2 = 4}
\]

Would you like more details or have any questions about this process?

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Here are 5 related questions you might explore:

1. How do you find the circumcenter of a triangle using coordinate geometry?
2. What is the significance of the circumcenter in different types of triangles?
3. Can the circumcenter be outside the triangle? If so, when does this occur?
4. How do you derive the general equation of a circle given three points?
5. What is the relationship between the circumradius and the sides of the triangle?

**Tip:** When finding the circumcenter, remember that the perpendicular bisectors of the sides of a triangle always intersect at a single point, regardless of the type of triangle.

Learn how to find the equation of the circle that circumscribes a triangle formed by specific lines in coordinate geometry. This detailed solution covers finding intersections, using the circumcenter formula, and deriving the circle's equation. Suitable for students studying coordinate geometry in Grades 10-12.

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