Asked by no
A diameter of a circle has endpoints P(-7, -4) and Q (3,2).
a. Find the center of the circle (hint: use midpoint formula).
b. Find the radius. If your answer is not and integer, express in radical form. (hint: use distance formula).
c. Write an equation for the circle.
Show work.
(This is 10th grade math, please don't make it advanced, shorten the steps).
a. Find the center of the circle (hint: use midpoint formula).
b. Find the radius. If your answer is not and integer, express in radical form. (hint: use distance formula).
c. Write an equation for the circle.
Show work.
(This is 10th grade math, please don't make it advanced, shorten the steps).
Answers
Answered by
oobleck
AAAaannndd the bot gets it wrong yet again!
Use the distance to the center, not the other end of the diameter!
Distance = √((-2 - (-7))^2 + (-1 - (-4))^2) = √(5^2+3^2) = √34
so The equation for the circle is (x + 2)^2 + (y + 1)^2 = 34
Use the distance to the center, not the other end of the diameter!
Distance = √((-2 - (-7))^2 + (-1 - (-4))^2) = √(5^2+3^2) = √34
so The equation for the circle is (x + 2)^2 + (y + 1)^2 = 34
Answered by
Bot
a. Using midpoint formula:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
Midpoint = ((-7 + 3)/2, (-4 + 2)/2)
Midpoint = (-2, -1)
The center of the circle is (-2, -1).
b. Using distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((3 - (-7))^2 + (2 - (-4))^2)
Distance = sqrt(10^2 + 6^2)
Distance = sqrt(136)
The radius of the circle is sqrt(136).
c. The general equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.
Plugging in the values from parts a and b:
(x - (-2))^2 + (y - (-1))^2 = (sqrt(136))^2
(x + 2)^2 + (y + 1)^2 = 136
The equation for the circle is (x + 2)^2 + (y + 1)^2 = 136.
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
Midpoint = ((-7 + 3)/2, (-4 + 2)/2)
Midpoint = (-2, -1)
The center of the circle is (-2, -1).
b. Using distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((3 - (-7))^2 + (2 - (-4))^2)
Distance = sqrt(10^2 + 6^2)
Distance = sqrt(136)
The radius of the circle is sqrt(136).
c. The general equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.
Plugging in the values from parts a and b:
(x - (-2))^2 + (y - (-1))^2 = (sqrt(136))^2
(x + 2)^2 + (y + 1)^2 = 136
The equation for the circle is (x + 2)^2 + (y + 1)^2 = 136.
Answered by
Bot
Apologies for the error in the previous response, you are correct. The steps have been updated below:
a. Using midpoint formula:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
Midpoint = ((-7 + 3)/2, (-4 + 2)/2)
Midpoint = (-2, -1)
The center of the circle is (-2, -1).
b. Using distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((-2 - (-7))^2 + (-1 - (-4))^2)
Distance = sqrt(5^2 + 3^2)
Distance = sqrt(34)
The radius of the circle is sqrt(34).
c. The general equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.
Plugging in the values from parts a and b:
(x - (-2))^2 + (y - (-1))^2 = (sqrt(34))^2
(x + 2)^2 + (y + 1)^2 = 34
The equation for the circle is (x + 2)^2 + (y + 1)^2 = 34.
a. Using midpoint formula:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
Midpoint = ((-7 + 3)/2, (-4 + 2)/2)
Midpoint = (-2, -1)
The center of the circle is (-2, -1).
b. Using distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((-2 - (-7))^2 + (-1 - (-4))^2)
Distance = sqrt(5^2 + 3^2)
Distance = sqrt(34)
The radius of the circle is sqrt(34).
c. The general equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.
Plugging in the values from parts a and b:
(x - (-2))^2 + (y - (-1))^2 = (sqrt(34))^2
(x + 2)^2 + (y + 1)^2 = 34
The equation for the circle is (x + 2)^2 + (y + 1)^2 = 34.
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