Assignment 19

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15:11:11 **** query 2.5.22 (was 2.4.18) Parallel to x - 2 y = -5 containing (0,0) **** Give your equation for the requested line and explain how you obtained it.

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RESPONSE --> x-2y = -5 containing (0,0) slope = 1 y - y1 = m(x - x1) y-0 = 2 (x - 0) y = 2x or 2x - y = 0

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15:11:56 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line parallel to this will therefore have slope 1/2. Point-slope form gives us y - 0 = 1/2 * (x - 0) or just y = 1/2 x. **

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RESPONSE --> I would have got it correct it i would have solved for y first.

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15:20:26 **** query 2.5.28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it.

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RESPONSE --> perp. x - 2y = -5 or y = x + 5 cont. (0,4) slope = 1 y - 4 = 1(x - 0) y - 4 = x y = x + 4

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15:20:40 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line perpendicular to this will therefore have slope -2/1 = -2. Point-slope form gives us y - 4 = -2 * (x - 0) or y = -2 x + 4. **

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RESPONSE --> I needed to use 1/2 not 1

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15:31:50 **** query 2.3.10 (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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RESPONSE --> (0,1) and (2,3) center = midpt. of (0,1) and (2,3) = 0+2 / 2 ; 1+3/2 = 2/2 , 4/2 = (1,2) radius = (1,2) to (2,3) = sqrt( (2-1)^2 + (3-2)^2 ) = sqrt (1^2 + 1^2) = sqrt (2) = 1.4 equation= (x-1)^2 + (y-2)^2 = 2

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15:32:07 ** The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = 2. This distance is a diameter so that the radius is 1/2 (2) = 1. The equation (x-h)^2 + (y-k)^2 = r^2 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). **

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RESPONSE --> correct.

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15:37:46 **** query 2.3.16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3?

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RESPONSE --> (h,k) = (1,0) r = 3 (x - 1)^2 + (y - 0)^2 = 3^2 (x-1)^2 + y^2 = 9 standard form: x^2 + y^2 - 9y + 9 = 9 x^2 + y^2 - 9y = 0

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15:38:22 ** 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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RESPONSE --> correct I think.????

You got the standard form (x-1)^2 + y^2 = 9.

However this is not equivalent to x^2 + y^2 - 9y + 9 = 9. It is equivalent to x^2 - 2x + y^2 = 8, as stated in the given solution.

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15:44:13 query 2.3.24 (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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RESPONSE --> x^2 + (y-1)^2 = 1 center = (0,0) radius = 1 If graphed, the point would be on the origin at (0,0) and a radius of 1 that would equal to the points (2,0) (0,2) (-2,0) and (0,-2) x-int. y = 0 x^2 +0 = 1 x^2 = 1 x = +- 1 (-1,0) , (1,0) y-int. x = 0 0 + y^2 = 1 y^2 = 1 y = +-1 (0, -1) , (0, 1)

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15:44:47 ** 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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RESPONSE --> I am not sure if I got this right or wrong.

Not completely right and not completely wrong.

The center is (0, -1) and r = sqrt(r^2) = sqrt(1) = 1.

You had the center being (0, 0).

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15:58:44 **** query 2.3.34 (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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RESPONSE --> 2x^2 + 2y^2 + 8x + 7= 0 (2x^2 + 8x) + (2y) = -7 (2x^2 + 8x + 16) + (2y +1) = -7 + 16 + 1 (2x^2 + 8x +16) + (2y + 1) = 10 radius = 10 but I cannot factor out the first one. I am having trouble or something.

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15:59:18 ** 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. From the standard form of the equation for a circle we see 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 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. **

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RESPONSE --> I didn't really understand that entire problem.

When you complete the square, the part about adding the square of 1/2 the coefficient of x works only when x^2 has coefficient 1.

Also there is no reason to complete the square on y, since there is only a y^2 term, with no y term.

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16:10:02 **** query 2.3.30 (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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RESPONSE --> diameter (4,3) and (0,1) center = 4+0/2 ; 3+1/2 = (2,2) radius = sqrt( (4-2)^2 + (2-3)^2 ) = sqrt (2^2 + (-1)^2 ) = sqrt (5) equation = (x - 2)^2 + (y-2)^2 = (sqrt(5))^2 x^2 - 4x +4 + y^2 - 4y + 4 = 5 x^2 + y^2 -4x - 4y +3 = 0

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16:10:14 ** 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 (x-2)^2 + (y-2)^2 = (sqrt(5))^2 or (x-2)^2 + (y-2)^2 = 5. **

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RESPONSE --> correct.

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Good. Let me know if you have questions.