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?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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

 

 

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 (if necessary):

 

 

 

 

 

 

 

 

 

 

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?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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?

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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 (h, k) = (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 follows that

(y-1) = +-1.

 

If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y = 0. 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?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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

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

 

Matching this with our equation

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

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)

 

 

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):

 

 

 

 

 

 

 

 

 

 

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.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

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

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. **

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating: