Standard form of the equation of a circle involving two points.
For parts (a), (b), and (c), consider the points $\mathrm A=(-1,3)$ and $\mathrm B=(5,3)$ in the $xy$-plane. Distance is measured in meters.
c) Find the standard form of the equation of a circle that has the line segment $\overline{\mathrm{AB}}$ as the diameter. (The circle has $\mathrm A$ and $\mathrm B$ as endpoints of the diameter)
With the following equations I arrived that $(3,0)$ is the center of the line: \begin{gather*} X = \frac{(-1 -5)} 2 = -3 \\ Y = \frac{(3 -3)} 2 = 0 \end{gather*} and then with the following I concluded that. $$(x-1)^2 + (y-1)^2 = \sqrt{13}$$
Can anyone verify this? I feel unease by my answer.
$\endgroup$ 21 Answer
$\begingroup$The center of the circle is the midpoint of the diameter: $(2, 3).$
The radius has length $3 m$ because the diameter has length $6 m$.
Then the equation for the circle is straightforward:
$$(x-2)^2 + (y-3)^2 = 9 m^2.$$
The $x$ coordinate of the center is $\frac{1}{2} \cdot (5 + (-1)) = 2 m.$
Actually, your $y$ coordinate was off, too: $\frac{1}{2} \cdot (3 + 3) = 3 m.$
You take the average of the coordinates to get the midpoint. This is an addition, not a subtraction.
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