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To find the equation of the circle with endpoints [tex]\(P = (-2, -1)\)[/tex] and [tex]\(Q = (4, 3)\)[/tex] as the diameter, we follow these steps:
1. Find the center of the circle: The center of the circle is the midpoint of the diameter PQ.
- Use the midpoint formula:
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
- Plug in the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex]:
[tex]\[ \left( \frac{-2 + 4}{2}, \frac{-1 + 3}{2} \right) = (1, 1) \][/tex]
Therefore, the center of the circle is [tex]\((1, 1)\)[/tex].
2. Find the radius of the circle: The radius is half the length of the diameter PQ.
- First, find the length of the diameter using the distance formula:
[tex]\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
- Plug in the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex]:
[tex]\[ \sqrt{(4 - (-2))^2 + (3 - (-1))^2} = \sqrt{(4 + 2)^2 + (3 + 1)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \][/tex]
- The radius is half of the diameter:
[tex]\[ \frac{\sqrt{52}}{2} = \sqrt{13} \][/tex]
3. Express the equation of the circle: The equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
- Here, [tex]\(h = 1\)[/tex], [tex]\(k = 1\)[/tex], and [tex]\(r^2 = 13\)[/tex].
Thus, substituting these values into the equation of the circle, we get:
[tex]\[ (x - 1)^2 + (y - 1)^2 = 13 \][/tex]
1. Find the center of the circle: The center of the circle is the midpoint of the diameter PQ.
- Use the midpoint formula:
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
- Plug in the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex]:
[tex]\[ \left( \frac{-2 + 4}{2}, \frac{-1 + 3}{2} \right) = (1, 1) \][/tex]
Therefore, the center of the circle is [tex]\((1, 1)\)[/tex].
2. Find the radius of the circle: The radius is half the length of the diameter PQ.
- First, find the length of the diameter using the distance formula:
[tex]\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
- Plug in the coordinates of [tex]\(P\)[/tex] and [tex]\(Q\)[/tex]:
[tex]\[ \sqrt{(4 - (-2))^2 + (3 - (-1))^2} = \sqrt{(4 + 2)^2 + (3 + 1)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \][/tex]
- The radius is half of the diameter:
[tex]\[ \frac{\sqrt{52}}{2} = \sqrt{13} \][/tex]
3. Express the equation of the circle: The equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
- Here, [tex]\(h = 1\)[/tex], [tex]\(k = 1\)[/tex], and [tex]\(r^2 = 13\)[/tex].
Thus, substituting these values into the equation of the circle, we get:
[tex]\[ (x - 1)^2 + (y - 1)^2 = 13 \][/tex]
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