Assignment 19

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course MTH 158

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?

Your solution:

In order to find the center f the circle we find the midpoint between these points.

M = 2 + 0 / 2, 3 + 1 / 2

M = 1, 2

So the Center is point 1,2

Next to find the radius we find the distance between either 0,1 and 1,2 or 2,3 and 1,2

D = sqrt (3-2)^2 + (2-1)^2

D = sqrt (1) + (1)

D = sqrt 2

So the radius is sqrt 2

The equation can be acquired by plugging in a point and the radius into the standard form

(x - 1)^2 + (y - 2)^2 = sqrt2^2

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

confidence rating #$&*: 2

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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

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?

Your solution:

Standard form is

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

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

The center will be equal to h,k so the center is (0,1).

Radius is 1 since 1^2 = 1

The graph is in both quadrant I and II

X intercept is 0, y intercept is 0 and 2

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)

Self-critique (if necessary):

I didn’t know to set x and y to zero when finding the intercepts. I will correct the error I made in future problems.

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Self-critique Rating: 3

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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?

Your solution:

First I reverse engineered this problem from general form to standard form and got

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

From this I got (-2,0) as the center and 0 as the radius

confidence rating #$&*:

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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. **

Self-critique (if necessary):

Again I failed to do the x and y intercepts for this problem. Also I got an incorrect answer for the radius. I think I made an error when converting this problem to standard form. I will review this and find the mistake

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Self-critique Rating: 2

@& 2 ( x - 2)^2 would give you 2 x^2 - 8 x, not the 2 x^2 + 8 x that occurs in the question.*@

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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.

Your solution:

First I found the midpoint between the two points of the diameter in order to acquire the center.

M = 4 + 0 / 2, 3 + 1 / 2

M = 2, 2

Now the distance between either point and the center must be found in order to acquire the radius.

D = sqrt (2 - 0)^2 + (2 - 1)^2

D = sqrt 5

Now just plug in a point with the radius and you have

(x - 2)^2 + (y - 2)^2 = sqrt5^2 or

(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