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

[mathdad]

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?

 

[MarkFL]

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?

 

[mathdad]

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.

 

[HallsofIvy]

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.

 

[mathdad]

(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]

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

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.

 

[mathdad]

What is the connection between my reply and the angry elephant picture?

 

[MarkFL]

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.

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.

 

[mathdad]

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.