To find the center of the circle that passes through the vertices of triangle ABC with vertices A(-2, 5), B(-2, -3), and C(2, -3), we can follow these steps:
### Step 1: Identify the vertices of the triangle
The vertices of the triangle are given as:
- A = (-2, 5)
- B = (-2, -3)
- C = (2, -3)
### Step 2: Determine the lengths of the sides of the triangle
We will calculate the lengths of the sides AB, BC, and CA using the distance formula:
\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
#### Calculate AB:
\[
AB = \sqrt{((-2) - (-2))^2 + ((-3) - 5)^2} = \sqrt{0^2 + (-8)^2} = \sqrt{64} = 8
\]
#### Calculate BC:
\[
BC = \sqrt{(2 - (-2))^2 + ((-3) - (-3))^2} = \sqrt{(2 + 2)^2 + 0^2} = \sqrt{4^2} = 4
\]
#### Calculate CA:
\[
CA = \sqrt{((-2) - 2)^2 + (5 - (-3))^2} = \sqrt{(-4)^2 + (8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}
\]
### Step 3: Check if triangle ABC is a right triangle
To check if triangle ABC is a right triangle, we can use the Pythagorean theorem. We need to check if \(AB^2 + BC^2 = CA^2\).
Calculating:
\[
AB^2 = 8^2 = 64
\]
\[
BC^2 = 4^2 = 16
\]
\[
CA^2 = (4\sqrt{5})^2 = 80
\]
Now check:
\[
AB^2 + BC^2 = 64 + 16 = 80 = CA^2
\]
Since this condition holds, triangle ABC is a right triangle with the right angle at B.
### Step 4: Find the circumcenter of triangle ABC
The circumcenter of a right triangle is located at the midpoint of the hypotenuse. The hypotenuse in this case is AC.
#### Calculate the midpoint of AC:
The midpoint M of a line segment with endpoints (x1, y1) and (x2, y2) is given by:
\[
M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
\]
Applying this to points A and C:
\[
M = \left(\frac{-2 + 2}{2}, \frac{5 + (-3)}{2}\right) = \left(\frac{0}{2}, \frac{2}{2}\right) = (0, 1)
\]
### Conclusion
The center of the circle that passes through the vertices of triangle ABC is at the point (0, 1).
### Final Answer
The center of the circle is: **(0, 1)**
