#$&* course Mth 158 7/8/12 at 11:27 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:
.............................................
Given Solution: * * The center of the circle is at the midpoint between the endpoints of the diameter, at x coordinate (0 + 2) / 2 = 1 and y coordinate (1 + 3) / 2 = 2. i.e., the center is at (1, 2). Using these coordinates, the general equation (x-h)^2 + (y-k)^2 = r^2 of a circle becomes (x-1)^2 + (y-2)^2 = r^2. Substituting the coordinates of the point (0, 1) we get (0-1)^2 + (1-2)^2 = r^2 so that r^2 = 2. Our equation is therefore (x-1)^2 + (y - 2)^2 = 2. You should double-check this solution by substituting the coordinates of the point (2, 3). Another way to find the equation is to simply find the radius from the given points: The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = sqrt(4 + 4) = sqrt(8) = 2 * sqrt(2). This distance is a diameter so that the radius is 1/2 (2 sqrt(2)) = sqrt(2). * The equation of a circle centered at (1, 2) and having radius sqrt(2) is (x-1)^2 + (y - 2)^2 = (sqrt(2)) ^ 2 or (x-1)^2 + (y - 2)^2 = 2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.4.14 / 16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We plug into the formula: (x-1)^2 + (y - 0)^2 = 3^2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You plug into the formula: (x - 0)^2 + (y - 1)^2 = 1. Your answers are: h = 0, k = 1, the center is (0, -1) and the radius is 1. To find the x intercept: you plug in 0 for y. x^2 + ( 0 -1)^2 = 1. So your x intercept is (0,0). To find the y intercept, you plug in 0 for x. 0 + (y -1)^2 =1. The y intercept is (0, -2). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example the equation can be written as (x - 0)^2 + (y-1)^2 = 1 So h = 0 and k = 1, and r^2 = 1. The center of the is therefore (0, -1) and r = sqrt(r^2) = sqrt(1) = 1. The x intercept occurs when y = 0: x^2 + (y-1)^2 = 1. I fy = 0 we get x^2 + (0-1)^2 = 1, which simplifies to x^2 +1=1, or x^2=0 so that x = 0. The x intercept is therefore (0, 0). The y intercept occurs when x = 0 so we obtain 0 + (y-1)^2 = 1, which is just (y - 1)^2 = 1. It follow that (y-1) = +-1. If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y -2. So the y-intercepts are (0,0) and (0,-2) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.4.32 / 34 (was 2.4.48). 2 x^2 + 2 y^2 + 8 x + 7 = 0 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have 2x^2 + 2y^2 + 8x + 7 = 0. We subtract 7 from both sides. This gives us 2x^2 + 2y^2 + 8x = -7. We divide by 2. We have x^2 + y^2 + 4x = -7/2. We work the x^2 part and get: x^2 + 4x + 4 + y^2 = -7/2 + 4. We factor: (x+2)^2 + y^2 = ½. We plug into our equation: (x + 2)^2 + y^2 = ½. Answers: h = -2, k = 0, r^2 = ½ and the center is (-2,0). To get the x intercept plug in y = 0, you have (x + 2)^2 + 0^2 = ½. We get x + 2 = +/-sqrt of ½. Your x intercept is -1.3 and -2.7. To find the y intercept, plug in 0 for x, you get (0 + 2)^2 + y^2 = ½. You have 4 + y^2 = ½. You subtract 4 from both sides, which gives you a negative answer. The y intercept cannot be negative so there isnt one. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ********************************************* 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, you plug in your answers: (4+0)/2, (3+1)/2 = (2,2). To find the radius: is sqrt ((2-0)^2 + (2-1)^2) = sqrt(5). You plug into the equation: (x-2)^2 + (y -2)^2 = sqrt 5^2. You will end up with (x-2)^2 + (y-2)^2 = 5. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!