#$&* course Mth 158 330 7/1 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:
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Given Solution: * * The center of the circle is at the midpoint between the endpoints of the diameter, at x coordinate (0 + 2) / 2 = 1 and y coordinate (1 + 3) / 2 = 2. i.e., the center is at (1, 2). Using these coordinates, the general equation (x-h)^2 + (y-k)^2 = r^2 of a circle becomes • (x-1)^2 + (y-2)^2 = r^2. Substituting the coordinates of the point (0, 1) we get (0-1)^2 + (1-2)^2 = r^2 so that r^2 = 2. Our equation is therefore • (x-1)^2 + (y - 2)^2 = 2. You should double-check this solution by substituting the coordinates of the point (2, 3). Another way to find the equation is to simply find the radius from the given points: The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = sqrt(4 + 4) = sqrt(8) = 2 * sqrt(2). This distance is a diameter so that the radius is 1/2 (2 sqrt(2)) = sqrt(2). * The equation of a circle centered at (1, 2) and having radius sqrt(2) is • (x-1)^2 + (y - 2)^2 = (sqrt(2)) ^ 2 or • (x-1)^2 + (y - 2)^2 = 2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 2.4.14 / 16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x-h)^2 + (y-k)^2=r^2 (x-1)^2 + (y-0)^2=3^2 Expand out then foil to: X^2-2x+1+y^2=9 Subtract 1 from each side: X^2-2x+y^2=8 confidence rating #$&*: 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): ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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 - h)^2 + (y - k)^2 = r^2 (x -0)^2 + (y-1)^2=1 So we know that the center is (0, 1) and that the radius is 1. Plot the points and find that the x intercepts include (0,0) and the y intercepts are at (0,0) and (0,2). 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 = 0. So the y-intercepts are (0,0) and (0,-2) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Why would the y intercept be zero and negative 2 instead of (0, 2)? I know that I did not use the formula but if you do graph it how could it intersect at -2 if the radius is 1?
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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): ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the center of the circle you plug in the numbers: (4+0)/2=2 and (3+1)/2=2 (2,2) is the center of the circle To find the distance from (2,2) and (0,1) you would do the following: Sqrt(2-0)^2 +(2-1)^2 which equals sqrt(5) You then plug that into the equation: (x-h)^2 + (y-k)^2=r^2 (x-2)^2 + (y-2)^2=sqrt(5)^2 Which equals (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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!