assignment 19

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course Mth 158

10/30 7

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:

starting with finding the center you take the 2 given coordinates which from each endpoints of the diameter, add the x's together and divide the result by 2 and add the y's together and divide the result by 2 and you get the coordinates of the center.

(0+2)/2=1 and (1+3)/2=2 so the center is (1,2)

now you can find the radius with the equation (x-h)^2 + (y-k)^2 = r^2 because you can plug in the points of the center to get (x-1)^2 + (y-2)^2 = r^2 and if you plug in the coordinates (0,1) from the diameter's end point you get (0-1)^2 + (1-2)^2 = r^2, then 1+1=r^2 and finally 2=r^2

the equation was (x-1)^2 + (y-2)^2 = r^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:

starting with equation of a circle you get (x-1)^2 +(y - 0)^2 = 3^2

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

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

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

x^2-2x+y^2=8 and that's about as far as I can go

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:

judging by x^2 + (y-1)^2 = 1, center points would be (0,1) and r=1

after graphing the circle by starting with the center coordinates and using the radius to find 4 points around the center, I got (1,1), (0,2), (-1,1), (0,0)

now start with the equation and let y=0 to find the x-int x^2 + (0-1)^2 = 1

x^2+(-1)^2=1, x^2+1=1, x^2=0 so x-int=(0,0)

to find y-int let x=0 so 0^2 + (y-1)^2 = 1, (y-1)^2=1, (y-1)=(+-)sqrt(1), y-1=(+-)1

y-1=1, y=2 or y-1= -1 , y=0 so y-int is (0,2) and (0,0)

the coordinates match what I graphed, so I assume they're right

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:

so to try to simplify the equation into something we can interpret to find the center, radius, etc we take 2 x^2 + 2 y^2 + 8 x + 7 = 0 and rewrite it as 2 x^2 + 8 x +2 y^2 + 7 = 0 or 2 x^2 + 8 x +2 y^2=-7 and then simplify it by dividing everything on both sides by 2 to get x^2+4x+y^2=-3.5 from here we can complete the square of x by adding 4 to both sides like x^2+4x+4+y^2=-3.5+4 and get

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

so now we can say that the center is (-2,0) and that the radius is sqrt(.5)

it's harder to just physically make a graph with that kind of result for a radius, so instead I just added and subtracted sqrt(.5) from the x and y coordinates to get (-2+sqrt(.5), 0), (-2, sqrt(.5)), (-2-sqrt(.5), 0), and (-2, -sqrt(.5))

x-int let y=0 so (x+2)^2+0^2=.5, (x+2)^2=.5, x+2=(+-)sqrt(.5), x=-2(+-)sqrt(.5) so x-int=(-2+sqrt(.5), 0) and (-2-sqrt(.5), 0)

y-int let x=0 so (0+2)^2+y^2=.5, 2^2+y^2=.5, 4+y^2=.5, y^2= -3.5 so y= (+-)sqrt(-3.5) so y-int would be (0,sqrt(-3.5)) and (0,-sqrt(-3.5))

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):

I just went ahead and used decimals, is it still right then?

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Self-critique Rating:ok

@& Fractions aren't subject to roundoff error the way decimals are.

1/2 = .5 and that is exact.

However 1/3 is .333...., which continues repeating forever. So, for example, writing .33 instead of 1/3 would involve roundoff error.

Sometimes this is acceptable, but it's difficult to judge. Unless a fraction can be represented exactly as a decimal, the fractional notation is generally preferable.*@

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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:

same as first question: add the x's together and divide by 2, and do the same to the y's so we get

(4+0)/2= 2 and (3+1)/2=2 so the midpoint of the diameter(and the center of the circle) would be (2,2)

from there we can say (x-2)^2+(y-2)^2=r^2 plug in one set of the coordinates given for the diameter and say (4-2)^2+(3-2)^2=r^2 -solve

4+1=r^2, 5=r^2, r= sqrt(5) so the final equation actually says

(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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assignment 19

&#This looks good. See my notes. Let me know if you have any questions. &#