Answer:
center (1,2)
radius: 3√2 units
Step-by-step explanation:
To find the center and radius of a circle given the endpoints of its diameter, follow these steps:
Find the midpoint of the diameter:
Use the midpoint formula to find the coordinates of the center of the circle, which will be the midpoint of the diameter.
Let the endpoints of the diameter be [tex]\sf (x_1, y_1) = (-2, -1) [/tex] and [tex]\sf (x_2, y_2) = (4, 5) [/tex].
The coordinates of the midpoint [tex]\sf (h, k) [/tex] can be found using:
[tex]\sf h = \dfrac{x_1 + x_2}{2} [/tex]
[tex]\sf k = \dfrac{y_1 + y_2}{2} [/tex]
Substituting the given points:
[tex]\sf h = \dfrac{-2 + 4}{2} = \dfrac{2}{2} = 1 [/tex]
[tex]\sf k = \dfrac{-1 + 5}{2} = \dfrac{4}{2} = 2 [/tex]
Therefore, the coordinates of the center [tex]\sf (h, k) [/tex] are [tex]\sf (1, 2) [/tex].
Find the radius of the circle:
Use the distance formula to find the radius of the circle, which is half of the length of the diameter.
The distance [tex]\sf d [/tex] between the endpoints [tex]\sf (x_1, y_1) [/tex] and [tex]\sf (x_2, y_2) [/tex] of the diameter is:
[tex]\sf d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
Substituting the given points:
[tex]\sf d = \sqrt{(4 - (-2))^2 + (5 - (-1))^2} [/tex]
[tex]\sf d = \sqrt{(4 + 2)^2 + (5 + 1)^2} [/tex]
[tex]\sf d = \sqrt{6^2 + 6^2} [/tex]
[tex]\sf d = \sqrt{36 + 36} [/tex]
[tex]\sf d = \sqrt{72} [/tex]
[tex]\sf d = 6\sqrt{2} [/tex]
Therefore, the radius [tex]\sf r [/tex] of the circle (which is half the diameter) is:
[tex]\sf r = \dfrac{d}{2} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2} [/tex]
Therefore, the center of the circle is [tex]\sf (1, 2) [/tex] and the radius is [tex]\sf \boxed{3\sqrt{2}} [/tex].
