Write the Standard Form of the Equation for this Circle

In summary, in this conversation, the speaker is trying to find the standard form of the equation of a circle with the given characteristics, which are the endpoints of a diameter. They discuss using the distance formula and the midpoint formula to find the radius and the center of the circle. Eventually, the speaker's friend solves the problem and provides the equation.
  • #1
nycmathguy
Homework Statement
Write standard form of the equation of a circle.
Relevant Equations
Equation of a circle not centered at the origin.

(x - h)^2 + (y - k)^2 = r^2
Chapter 1, Section 1.2

Write the standard form of the equation of the circle with the given characteristics.

74. Endpoints of a diameter: (11, −5), (3, 15)

I want to know if the following steps are correct for me to answer the above question.

Steps:

1. Find the distance between the points.

2. Divide the distance by 2 to find the radius.

3. Plug into (x - h)^2 + (y - k)^2 = r^2

You say?
 
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  • #2
nycmathguy said:
Homework Statement:: Write standard form of the equation of a circle.
Relevant Equations:: Equation of a circle not centered at the origin.

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

Steps:

1. Find the distance between the points.

2. Divide the distance by 2 to find the radius.

3. Plug into (x - h)^2 + (y - k)^2 = r^2
Looks reasonable so far, but how do you use those two endpoints to find h and k? :smile:
 
  • #3
Do you know the midpoint formula?
 
  • #4
cbarker1 said:
Do you know the midpoint formula?

Yes, I know the midpoint formula. What about it?
 
  • #5
berkeman said:
Looks reasonable so far, but how do you use those two endpoints to find h and k? :smile:
I would plug the x and y values of the endpoints in the standard form of an equation of a circle to find h and k. Yes?
 
  • #6
nycmathguy said:
I would plug the x and y values of the endpoints in the standard form of an equation of a circle to find h and k. Yes?
Sorry, I have no idea what that means. Instead, I would follow this hint:
cbarker1 said:
Do you know the midpoint formula?
 
  • #7
nycmathguy said:
You say?
I say that once again, you are not following through.

Draw the two points on a graph. Calculate the midpoint numerically and see if it looks right on the graph. When you have the midpoint, write the equation of the circle. There's more for a full follow-through.
 
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  • #8
nycmathguy said:
Yes, I know the midpoint formula. What about it?
How is the mid-point of a diameter of a circle related to the location of the circle's center ?
 
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  • #9
nycmathguy said:
Yes, I know the midpoint formula. What about it?
So what is the formula, then?
 
  • #10
SammyS said:
How is the mid-point of a diameter of a circle related to the location of the circle's center ?

The midpoint of a circle divides the diameter into two radii.
 
  • #11
nycmathguy said:
The midpoint of a circle divides the diameter into two radii.
I asked about the mid-point of a diameter of a circle. How is the mid-point related to the center of the circle?
 
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  • #12
nycmathguy said:
The midpoint of a circle divides the diameter into two radii.
Quite futzin' around just DRAW it and get on with solving the problem. I don't understand why you have such an issue with follow through. You are going to keep having trouble with these problems if you don't get past that issue.
 
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  • #13
phinds said:
Quite futzin' around just DRAW it and get on with solving the problem. I don't understand why you have such an issue with follow through. You are going to keep having trouble with these problems if you don't get past that issue.
Follow through in what way? I'm stuck here.
 
  • #14
SammyS said:
I asked about the mid-point of a diameter of a circle. How is the mid-point related to the center of the circle?

Midpoint = [(11 + 3)/2, (-5 +15)/2]

Midpoint = (14/2, 10/2)

Midpoint = (7, 5)

This point = (h, k) = (7, 5).

So, h = 7 and k = 5.

I now plug into (x - h)^2 + (y - k)^2 = r^2.

Wait, I need the radius.

To find the radius, I must use the distance formula points.

So far, so good, yes?
 
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  • #15
phinds said:
Quite futzin' around just DRAW it and get on with solving the problem. I don't understand why you have such an issue with follow through. You are going to keep having trouble with these problems if you don't get past that issue.

I found h to 7 and k to be 5.

My friend completed the problem.

Here is her work:

distance between center (7,5) and point (3, 15) is the length of radius:

r=sqrt((7-3)^2+(5-15)^2)
r=sqrt(4^2+(-10)^2)
r=sqrt(16+100)
r=sqrt(116)

then r^2=116
(x - 7)^2 + (y - 5)^2 = 116=> your equation
 
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  • #16
nycmathguy said:
Follow through in what way? I'm stuck here.
Do you not understand how to put points on a graph?
 
  • #17
phinds said:
Do you not understand how to put points on a graph?
Forget it. Problem has been solved.
 
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What is the standard form of an equation for a circle?

The standard form of an equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the center of the circle and r represents the radius.

How do I determine the center and radius of a circle from its standard form equation?

The center of the circle is represented by the values of h and k in the equation, while the radius is the square root of r^2. For example, if the equation is (x - 2)^2 + (y + 3)^2 = 9, the center is (2,-3) and the radius is 3.

Can the standard form of a circle's equation have negative values for h and k?

Yes, the center of a circle can have negative values for h and k. This simply means that the circle is not centered at the origin (0,0) but rather at a point with coordinates (h,k).

Can the radius of a circle be a negative number?

No, the radius of a circle cannot be a negative number. It represents the distance from the center to any point on the circle, so it must be a positive number.

How do I graph a circle using its standard form equation?

To graph a circle using its standard form equation, first determine the center and radius. Then, plot the center point on a coordinate plane and use the radius to draw a circle around it. You can also plot a few points on the circle to ensure accuracy.

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