Assign 19

course Mth 158

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

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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?

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Your solution:

To find the center of the circle, first you have to find the midpoint of the given line

(0 + 2) / 2 = 1 and ( (1 + 3) / 2 = 2 this gives us the coordinate (1, 2) which is the center of the circle.

To find the radius of the circle you use the formula

(x – h)^2 + (y – k ) = r^2

(0 – 1)^2 + (1 – 2)^2 = r^2

(-1)^2 + (-1)^2 = r^2

2 = r^2 (When I get to this point, I am not sure if it’s 2 = r^2 or r = 2^2 Would the radius then be 2?)

The equation would be

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

confidence rating: 1

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Given Solution:

* *

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

• (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).

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

• (x-1)^2 + (y - 2)^2 = (sqrt(2)) ^ 2 or

• (x-1)^2 + (y - 2)^2 = 2

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Self-critique (if necessary):

There are no discrepancies

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Self-critique Rating: 3

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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?

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Your solution:

center (1, 0)

radius = 3

using the formula

(x-h)^2 + (y-k)^2 = r^2 we substitute h=1 k=0 r=3

(x-1)^2 + (y-0)^2 = 9 this is the standard form of the equation of a circle

confidence rating: 3

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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.

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Self-critique (if necessary):

There are no discrepancies

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Self-critique Rating: 3

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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?

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Your solution:

x^2 + (y-1)^2 = 1 write in standard form

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

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

Now we have

h = 0

k = 1

r^2 = 1 center (0,1) radius = 1

To get x intercept let y = 0

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

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

x^2 +1 = 1

x^2 = 0

x = 0

To get y intercept let x =0

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

y = +- 1

y -1 = 1 then y = 2

y + 1 = 1 y = 0

This gives us

(0,0) (0, -2)

confidence rating: 2

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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)

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Self-critique (if necessary):

There are no discrepancies

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Self-critique Rating:

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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?

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Your solution:

2x^2 + 2y^2 + 8x + 7 = 0 first group like terms

(2x^2 + 8x) + 2y^2 = -7 I know you divide by a common factor to get rid of parenthesis so I think the next step is this:

½(2x^2 + 8x) + ½(2y^2) = ½(-7)

x^2 + 4x + y^2 = -7/2 then you complete the square on both sides of the equation

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

(x +2)^2 + y^2 = ½ and then I get lost from here.

confidence rating: 0

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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. **

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Self-critique (if necessary):

I can see how you solved the problem, but remembering that many steps is going to take practice. I thought I knew how to do it until I started solving it.

Basically you complete the squares then factor. However there are a number of details.

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Self-critique Rating: 3

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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.

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Your solution:

Find the center of the circle by finding the midpoint of the two points.

(4+0)/2 = 2 (3+1)/2 = 2 the center of the circle is (2, 2)

The distance from (2, 2) to (0, 1) is (2-0)^2 + (2-1)^2 = 5

I am really having trouble with this section. I may need to stop and go back and read this chapter again.

confidence rating: 0

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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. **

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Self-critique (if necessary):

I can see how you solved the rest of the problem, but I am not confident that I could do it.

You had a good start. You just need to know the standard form of the equation of a circle which is

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

where the center is the point (h, k) and the radius is r.

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Self-critique Rating: 2

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&#Good responses. See my notes and let me know if you have questions. &#