Answer
Non esiste una circonferenza che passa per tutti e tre i punti dati A(-4, 0), B(-2, -1) e C(2, -3).
Solution
Okay, I need to determine if there exists an equation for a circle that passes through the points A(-4, 0), B(-2, -1), and C(2, -3). Let's break this down step by step.
First, I recall that the general equation of a circle in the Cartesian plane is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is its radius.
Given three points, I can set up a system of equations to solve for \(h\), \(k\), and \(r\). Plugging in the coordinates of points A, B, and C into the general equation:
1. For point A(-4, 0):
\[
(-4 - h)^2 + (0 - k)^2 = r^2 \quad \Rightarrow \quad (4 + h)^2 + k^2 = r^2 \quad \text{(Equation 1)}
\]
2. For point B(-2, -1):
\[
(-2 - h)^2 + (-1 - k)^2 = r^2 \quad \Rightarrow \quad (2 + h)^2 + (1 + k)^2 = r^2 \quad \text{(Equation 2)}
\]
3. For point C(2, -3):
\[
(2 - h)^2 + (-3 - k)^2 = r^2 \quad \Rightarrow \quad (2 - h)^2 + (3 + k)^2 = r^2 \quad \text{(Equation 3)}
\]
Now, I have three equations with three unknowns (\(h\), \(k\), and \(r\)). To solve for these, I can subtract Equation 2 from Equation 1 to eliminate \(r^2\):
\[
(4 + h)^2 + k^2 - [(2 + h)^2 + (1 + k)^2] = 0
\]
Expanding the squares:
\[
(16 + 8h + h^2) + k^2 - (4 + 4h + h^2 + 1 + 2k + k^2) = 0
\]
Simplifying:
\[
16 + 8h + h^2 + k^2 - 4 - 4h - h^2 - 1 - 2k - k^2 = 0
\]
\[
(16 - 4 - 1) + (8h - 4h) + (h^2 - h^2) + (k^2 - k^2) - 2k = 0
\]
\[
11 + 4h - 2k = 0 \quad \Rightarrow \quad 4h - 2k = -11 \quad \text{(Equation 4)}
\]
Next, subtract Equation 3 from Equation 2:
\[
(2 + h)^2 + (1 + k)^2 - [(2 - h)^2 + (3 + k)^2] = 0
\]
Expanding the squares:
\[
(4 + 4h + h^2) + (1 + 2k + k^2) - (4 - 4h + h^2 + 9 + 6k + k^2) = 0
\]
Simplifying:
\[
4 + 4h + h^2 + 1 + 2k + k^2 - 4 + 4h - h^2 - 9 - 6k - k^2 = 0
\]
\[
(4 + 1 - 4 - 9) + (4h + 4h) + (h^2 - h^2) + (k^2 - k^2) + (2k - 6k) = 0
\]
\[
-8 + 8h - 4k = 0 \quad \Rightarrow \quad 8h - 4k = 8 \quad \text{(Equation 5)}
\]
Now, I have two equations with two unknowns (\(h\) and \(k\)):
\[
\begin{cases}
4h - 2k = -11 \quad \text{(Equation 4)} \\
8h - 4k = 8 \quad \text{(Equation 5)}
\end{cases}
\]
To solve for \(h\) and \(k\), I can use the method of elimination. Multiply Equation 4 by 2 to align the coefficients of \(k\):
\[
8h - 4k = -22 \quad \text{(Equation 6)}
\]
Now, subtract Equation 5 from Equation 6:
\[
(8h - 4k) - (8h - 4k) = -22 - 8
\]
\[
0 = -30
\]
This is a contradiction, which means there is no solution for \(h\) and \(k\) that satisfies both equations simultaneously. Therefore, there is no circle that passes through all three given points A(-4, 0), B(-2, -1), and C(2, -3).
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