MHB How to Determine the Equation of a Circle from Given Diameter Coordinates?

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Determine the equation of the circle in standard form, given the coordinate of the diameter PQ.

P(-4, -2) and Q(6, 4)

Midpoint is (5, 3).

d = sqrt{(6-(-4))^2 + (4-(-2))

d = sqrt{(10)^2 + (6)^2}

d = sqrt{100 + 36}

d = r = sqrt{136}

Let d = distance = radius

(x - h)^2 + (y - k)^2 = r^2

(x - 5)^2 + (y - 3)^2 = [sqrt{136}]^2

(x - 5)^2 + (y - 3)^2 = 136

Correct?
 
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RTCNTC said:
Determine the equation of the circle in standard form, given the coordinate of the diameter PQ.

P(-4, -2) and Q(6, 4)

Midpoint is (5, 3).

d = sqrt{(6-(-4))^2 + (4-(-2))

d = sqrt{(10)^2 + (6)^2}

d = sqrt{100 + 36}

d = r = sqrt{136}

Let d = distance = radius

(x - h)^2 + (y - k)^2 = r^2

(x - 5)^2 + (y - 3)^2 = [sqrt{136}]^2

(x - 5)^2 + (y - 3)^2 = 136

Correct?

Let's graph your circle and the points P and Q as a check:

View attachment 7481

Hmmm...looks like the radius and center points aren't correct.

Can you spot the errors you made in applying the mid-point formula, and in determining the length of the radius?
 

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MarkFL said:
Let's graph your circle and the points P and Q as a check:
Hmmm...looks like the radius and center points aren't correct.

Can you spot the errors you made in applying the mid-point formula, and in determining the length of the radius?

On my way to work now. I will work out the math later this evening. I like the geometric picture. I would like to see more like it when answering questions involving algebra and geometry.
 
RTCNTC said:
Determine the equation of the circle in standard form, given the coordinate of the diameter PQ.

P(-4, -2) and Q(6, 4)
So you are given the coordinates of the endpoints of the diameter PQ.
Midpoint is (5, 3).
No, the midpoint is ((-4+ 6)/2, (-2+ 4)/2)= (1, 1).

d = sqrt{(6-(-4))^2 + (4-(-2))

d = sqrt{(10)^2 + (6)^2}

d = sqrt{100 + 36}

d = r = sqrt{136}
No, the diameter is not the radius! The radius is half the diameter, \sqrt{136}/2= \sqrt{34}

Let d = distance = radius

(x - h)^2 + (y - k)^2 = r^2

(x - 5)^2 + (y - 3)^2 = [sqrt{136}]^2

(x - 5)^2 + (y - 3)^2 = 136

Correct?
No, it is not.
 
(x - h)^2 + (y - k)^2 = r^2

(x - 1)^2 + (y - 1)^2 = sqrt{136}/2

(x - 1)^2 + (y - 1)^2 = [sqrt{34}]^2

(x - 1)^2 + (y - 1)^2 = 34

Correct?
 
MarkFL said:
...Can you spot the errors you made in applying the mid-point formula, and in determining the length of the radius?

HallsofIvy said:
So you are given the coordinates of the endpoints of the diameter PQ.

No, the midpoint is ((-4+ 6)/2, (-2+ 4)/2)= (1, 1).

No, the diameter is not the radius! The radius is half the diameter, \sqrt{136}/2= \sqrt{34}

No, it is not.

Jumanji-3.jpg
 
What is the connection between my reply and the angry elephant picture?
 
RTCNTC said:
What is the connection between my reply and the angry elephant picture?

I asked you to spot the errors you made, and then another person came along and obliviously answered the entire problem...this is called "trampling" and is against MHB's rules.

MHB Rule #14 said:
Do not explain hints prematurely. When you see someone already helping a member and waiting for the original poster to give feedback, do not give more hints or a full solution. Wait at least 24 hours for the original poster to respond.

Thus, I posted a picture of an elephant trampling a car as a humorous way to object to having my help trampled.
 
MarkFL said:
I asked you to spot the errors you made, and then another person came along and obliviously answered the entire problem...this is called "trampling" and is against MHB's rules.
Thus, I posted a picture of an elephant trampling a car as a humorous way to object to having my help trampled.
Thanks for explaining the picture to me. I will continue to post from the David Cohen precalculus textbook until I receive my Michael Kelly precalculus Quick Review book in about two weeks.
 

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