Assignment 19 24

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course Mth 158

3/8/13 Around 9:00PM

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:

 C = (1 , 2); r = sqrt(2); r^2 = 2

 Find M between two points for C:

 (0 + 2) / 2 = 1

(1 + 3) / 2 = 2

 C = (1 , 2)

 Use the distance formula to get distance between center and point for r:

 sqrt[(0 + 1)^2 + (1 - 2)^2] = r

 sqrt(1 + 1) = r

 sqrt(2) = r

 r^2 = 2

 Equation of circle:

 (x - 1)^2 + (y - 2)^2 = 2

 

 

confidence rating #$&*: 3

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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): OK

 

 

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Self-critique Rating: OK

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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:

 (x - 1)^2 + y^2 = 9

 

 

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): OK

 

 

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Self-critique Rating: OK

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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:

 C = (0 , 1); r = 1; the graph is in the first and second quadrants.

Find x-intercept: 

x^2 + 1 = 1

 x^2 = 0

 x-int = (0 , 0)

 Find y-intercept:

 (y - 1)^2 = 1

 y - 1 = 1

 y = 2

or

y - 1 = -1

y = 0

 y-int = (0 , 2) and (0 , 0)

 

 

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 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): OK

 

 

 

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Self-critique Rating: OK

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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

 2x^2 + 2y^2 + 8x = -7

 Divide both sides by 2:

 x^2 + y^2 + 4x = -7/2

Factor by grouping:

 (x^2 + 4x) + y^2 = -7/2

(x^2 + 4x + 4) + y^2 = -7/2 + 4

 (x + 2)^2 + y^2 = 1/2

 C = (2 , 0); r = sqrt(1/2) = sqrt(1) / sqrt (2) = 1 / sqrt(2) = 1 * sqrt(2) / sqrt(2)^2 = sqrt(2) / 2

Find x-intercept(s):

 (x + 2)^2 = 1/2

 x + 2 = sqrt(2) / 2

 x = (sqrt(2)) / (2) - 2

 x = (sqrt(2)) / (2) - 4/2

 x = (-4 + sqrt(2)) / 2

 or

 x = (-4 - sqrt(2)) / 2

 Find y-intercepts:

 4 + y^2 = 1/2

 y^2 = -7/2

 There are no y-int because a squared number cannot be negative.

 

 

confidence rating #$&*: 3

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Given Solution:

* *  We first want to complete the squares on the x and y terms:

 

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.

 

We interpret our result:

 

The standard form of the equation of a circle is

• (x - h)^2 + (y - k)^2 = r^2,

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

 

Matching this with our equation

• (x+2)^2 + y^2 = 1/2

we find that h = -2, k = 0 and r^2 = 1/2.  We conclude 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

• (note:  The above solutions are approximate.  The exact solutions can be expressed according to the following:

• sqrt(1/2) = 1 / sqrt(2) = sqrt(2) / 2, found by rationalizing the denominator; so sqrt(1/2) - 2 = sqrt(2)/2 - 2 = (sqrt(2) - 4) / 2.  This is an exact solution for one x intercept.  The other is (-sqrt(2) - 4) / 2.

 

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.  This is consistent with the fact that a circle centered at (2, 0) with radius sqrt(1/2) lies entirely to the right of the y axis. **

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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 M of given points (4, 3) and (0, 1):

 (4 + 0) / 2 and (3 + 1) / 2

 M = (2 , 2)

 Use distance formula for distance between M and (4, 3):

 sqrt[(4 - 2)^2 + (3 - 2)^2]

 sqrt[5]

 r = sqrt(5)

 r^2 = 5

 (x + 2)^2 + (y + 2)^2 = 5

 

 

confidence rating #$&*: 3

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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 found from the standard equation, which is

• (x - h)^2 + (y - k)^2 = r^2,

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

 

Since the center is at (2, 2) and the radius is sqrt(5), our equation is

 

(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): OK

 

 

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Self-critique Rating: OK

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