Assignment 19

course MTH 158

7/20/09 11:52 P.M.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

019. `* 19

*********************************************

Question: * 2.4.10 (was 2.4.30). (0,1) and (2,3) on diameter **** What are the center, radius and equation of the indicated circle?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

sqrt((2-0)^2 + (3-1)^2) = R

sqrt((2)^2 + (2)^2 = R

Sqrt(4 + 4) = R

sqrt(8) = R

R = 2* sqrt(2)

r ~ 2.83

Center of circle would be (0,1)

The equation would be

(2-0)^2 + (3-1)^2 = 2*sqrt(2) ~ 2.83

Confidence rating: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = 2. This distance is a diameter so that the radius is 1/2 (2) = 1.

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

(x-1)^2 + (y-2)^2 = r^2.

Substituting the coordinates of the point (0, 1) we get

(0-1)^2 + (1-2)^2 = r^2 so that

r^2 = 2.

Our equation is therefore

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

You should double-check this solution by substituting the coordinates of the point (2, 3). **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Not really sure where I went wrong. I cant figure out how you come up with sqrt( (2-0)^2 + (3-1)^2 ) = 2

The distance you found was the diameter, not the radius.

However the given solution did have an error in finding the radius at the beginning. That didn't affect the final answer in the given solution because that erroneous radius was never used.

Here's a modification that shows two different ways to find the equation:

The center of the circle is at the

midpoint between the endpoints of the diameter, at x coordinate (0 + 2) / 2 = 1

and y coordinate (1 + 3) / 2 = 2.  i.e., the center is at (1, 2).

 

Using these coordinates, the

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

of a circle becomes

 

 

Substituting the coordinates of the point (0, 1) we get

 

(0-1)^2 + (1-2)^2 = r^2 so that

r^2 = 2.

 

Our equation is therefore

 

 

You should double-check this solution by substituting the

coordinates of the point (2, 3).

 

 

 

 

Another way to find the equation is to

simply find the radius from the given points:

The distance from (0,1) to (2,3) is sqrt( (2-0)^2 +

(3-1)^2 ) = sqrt(4 + 4) = sqrt(8) = 2

* sqrt(2).

 

This distance is a

diameter so that the radius is 1/2 (2 sqrt(2)) =

sqrt(2).

*

 

The equation of a circle centered at

(1, 2) and having radius sqrt(2) is

 

:

Self-critique Rating:

*********************************************

Question: * 2.4.14 / 16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(x-1)^2 + (y-0)^2 = 3^2

Confidence rating: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

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

In this example we have (h, k) = (1, 0). We therefore have

(x-1)^2 +(y - 0)^2 = 3^2.

This is the requested standard form.

This form can be expanded and simplified to a general quadratic form. Expanding (x-1)^2 and squaring the 3 we get

x^2 - 2x +1+y^2 = 9

x^2 - 2x + y^2 = 8.

However this is not the standard form.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique Rating:

*********************************************

Question: * 2.4.22 / 24 (was 2.4.40). x^2 + (y-1)^2 = 1 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Standard form would be

(x-0)^2 + (y-1)^2 = 1

So (0,1) would be the center of the circle

The radius would be (1)

y intercept = 0 and 2

x intercept = 1 and -1

Set x to (0) to obtain the y intercept

(y-1)^2 = 1

sqrt(y-1)= sqrt(1)

y-1 = +/- 1

y= 1 +1

y = 2 and y = 1 - 1 y= 0

Set y to (0) to obtain the x intercept

(x-0)^2 = 1

sqrt (x-0) = sqrt(1)

x-0 = +/- 1

x = 1 and - 1

Confidence rating: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

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

In this example the equation can be written as

(x - 0)^2 + (y-1)^2 = 1

So h = 0 and k = 1, and r^2 = 1. The center of the is therefore (0, -1) and r = sqrt(r^2) = sqrt(1) = 1.

The x intercept occurs when y = 0:

x^2 + (y-1)^2 = 1. I fy = 0 we get

x^2 + (0-1)^2 = 1, which simplifies to

x^2 +1=1, or

x^2=0 so that x = 0. The x intercept is therefore (0, 0).

The y intercept occurs when x = 0 so we obtain

0 + (y-1)^2 = 1, which is just (y - 1)^2 = 1. It follow that

(y-1) = +-1.

If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y -2. So the y-intercepts are

(0,0) and (0,-2)

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique Rating:

*********************************************

Question: * 2.4.32 / 34 (was 2.4.48). 2 x^2 + 2 y^2 + 8 x + 7 = 0 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First factor this form

x^2 + 4x + 4 + y^2 = -7/2 +8/2

x^2 + 4x + 4 + y^2 = 1/2

(x+2)^2 + y^2 = 1/2

x+2)^2 + (y-0)^2 = 1/2

Center is (-2,0)

Radius is the sqrt(1/2) or .71

To find x and y intercepts set x to (0) to find y and Y to (0) to find x

(x+2)^2 = 1/2

sqrt both sides

x+2 = +/- sqrt(1/2)

x= -2(+/-)sqrt(1/2)

x = -1.29 and -2.71

Finding y

4 + y^2 = 1/2

y^2 = 1/2 - 8/2

y^2 = -7/2

no Y intercept because you cant take the square of a negative number

Confidence rating: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * Starting with

2x^2+ 2y^2 +8x+7=0 we group x and y terms to get

2x^2 +8x +2y^2 =-7. We then divide by the common factor 2 to get

x^2 +4x + y^2 = -7/2. We complete the square on x^2 + 4x, adding to both sides (4/2)^2 = 4, to get

x^2 + 4x + 4 + y^2 = -7/2 + 4. We factor the expression x^2 + 4x + 4 to obtain

(x+2)^2 + y^2 = 1/2. From the standard form of the equation for a circle we see that

the center is (-2,0)

the radius is sqrt (1/2).

To get the intercepts:

We use (x+2)^2 + y^2 = 1/2

If y = 0 then we have

(x+2)^2 + 0^2 = 1/2

(x+2)^2 = 1/2

(x+2) = +- sqrt(1/2)

x + 2 = sqrt(1/2) yields x = sqrt(1/2) - 2 = -1.3 approx.

x + 2 = -sqrt(1/2) yields x = -sqrt(1/2) - 2 = -2.7 approx

If x = 0 we have

(0+2)^2 + y^2 = 1/2

4 + y^2 = 1/2

y^2 = 1/2 - 4 = -7/2.

y^2 cannot be negative so there is no y intercept. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique Rating:

*********************************************

Question: * 2.4.40 / 30 (was 2.4.54). General equation if diameter contains (4, 3) and (0, 1). **** Give the general equation for your circle and explain how it was obtained.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I couldnt figure out how to get the radius from the diameter until I looked at your answer and it made sense then

Confidence rating: 0

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * The center of the circle is the midpoint between the two points, which is ((4+0)/2, (3+1)/2) = (2, 2).

The radius of the circle is the distance from the center to either of the given points. The distance from (2, 2) to (0, 1) is sqrt( (2-0)^2 + (2-1)^2 ) = sqrt(5).

The equation of the circle is therefore

(x-2)^2 + (y-2)^2 = (sqrt(5))^2 or

(x-2)^2 + (y-2)^2 = 5. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

After I read your solution I was able to figure this equation out. I will study this more before I take the test.

Self-critique Rating:

"

&#Very good work. Let me know if you have questions. &#