Geometry B Unit 7

TQ5) The center of the circle is at the point (-4, 6.75)(4, 8)(4, 9.25)(4, 16), and its radius is 1.252.53.756.25 units. The equation of this circle in standard form is (x - 4)^2 + (y - 8)^2 = 2.5(x + 4)^2 + (y + 8)^2 = 2.5(x - 4)^2 + (y - 8)^2 = 6.25(x + 4)^2 + (y + 8)^2 = 6.25.

(4, 8) 2.5 (x-4) ^2 + (y-8) ^2 =6.25

What is the equation of a circle with center (-3, -1) that contains the point (1, 2)?

(x + 3)^2 + (y + 1)^2 = 25

What is the standard form of this equation of a circle?x2 + y2 + 14x − 4y − 28 = 0

(x + 7)^2 + (y − 2)^2 = 81

What is the equation of a circle with center (2, -5) and radius 4?

(x − 2)^2 + (y + 5)^2 = 16

TQ2) Match the circle equations in general form with their corresponding equations in standard form.

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Move the sliders to set the values of r, h, and k, and record at least three different sets of data. Also record the equations of the corresponding circles.

1) S1: -2 S2: 4 S3: 3 S4: (x - (-2))^2 + (y - 4)^2 = 3^2 2) S1: -2 S2: 4 S3: 2 S4 (x - (-2))^2 + (y - 4)^2 = 2^2 3) S1: 2 S2: 4 S3: 3 S4: (x - 2)^2 + (y - 4)^2 = 3^2 4) S1: 2 S2: -2 S3: 3 S4: (x - 2)^2 + (y - (-2))^2 = 3^2 5) S1: 0 S2: -2 S3: 1 S4: (x - 0)^2+ (y - (-2))^2 = 1^2 6) S1: 3 S2: 0 S3: 1 S4: (x - 3)^2 + (y - 0)^2 = 1^2 Edmentum answer: ) S1: -2 S2: 4 S3: 3 S4: (x - (-2))^2 + (y - 4)^2 = 3^2 2) S1: -2 S2: 4 S3: 2 S4 (x - (-2))^2 + (y - 4)^2 = 2^2 3) S1: 2 S2: 4 S3: 3 S4: (x - 2)^2 + (y - 4)^2 = 3^2 4) S1: 2 S2: -2 S3: 3 S4: (x - 2)^2 + (y - (-2))^2 = 3^2 5) S1: 0 S2: -2 S3: 1 S4: (x - 0)^2+ (y - (-2))^2 = 1^2 6) S1: 3 S2: 0 S3: 1 S4: (x - 3)^2 + (y - 0)^2 = 1^2

How does the equation change when the radius changes? Unlike h and k, why is r always positive?

At the point when it changed , the steady term (r2) in the condition transforms—it increases as the span expands. The term r is consistently positive since it signifies the length of the span, which is a separation and not a position or area. Edmentum answer:When the radius is changed, the constant term (r2) in the equation changes—it becomes bigger as the radius becomes bigger. The term r is always positive because it denotes the length of the radius, which is a distance and not a position or location.

The equation of a circle in general form is Ax2 + By2 + Cx + Dy + E = 0, where A = B ≠ 0. How do the coefficients C and D change as the center of the circle crosses over the x- and y-axes? How do the coefficients C and D change if the radius of the circle changes? Experiment with different circles using the general form of the equation in GeoGebra, if you wish.

In the event that the focal point of the hover is to one side of the y-axis, C is negative and its total worth increments as the inside moves to one side. In the event that the inside is to one side of the y-axis, C is sure and its worth abatements as the middle moves to one side. D doesn't change when the middle moves evenly. On the off chance that the focal point of the hover is over the x-axis, D is negative and its supreme worth increments as the inside climbs. On the off chance that the middle is beneath the y-axis, D is certain and its worth reductions as the inside climbs. C doesn't change when the inside moves vertically. The estimations of C and D are not influenced when the sweep changes, as long as the inside remains the equivalent. Edmentum answer: If the center of the circle is to the right of the y-axis, C is negative and its absolute value increases as the center moves to the right. If the center is to the left of the y-axis, C is positive and its value decreases as the center moves to the right. D does not change when the center moves horizontally.If the center of the circle is above the x-axis, D is negative and its absolute value increases as the center moves up. If the center is below the y-axis, D is positive and its value decreases as the center moves up. C does not change when the center moves vertically.The values of C and D are not affected when the radius changes, as long as the center stays the same.

Move the center of the circle horizontally to the left and then to the right of the y-axis. How does the equation of the circle change as the center crosses the y-axis?

The variable h changes as the focal point of the circle moves evenly. The indication of h is negative when the middle is to one side of the y-axis and positive when it is to one side of the y-axis. The indication of the variable h flips when the inside moves over the y-axis. Edmentum answer: The variable h changes as the center of the circle moves horizontally. The sign of h is negative when the center is to the left of the y-axis and positive when it is to the right of the y-axis. The sign of the variable h flips when the center moves across the y-axis.

Move the center of the circle vertically so it lies above and then below the x-axis. How does the equation of the circle change as the center crosses the x-axis?

The variable k changes as the focal point of the circle moves vertically. The indication of k is sure when the middle is beneath the x-axis and negative when it is over the x-axis. The indication of the variable k flips when the inside moves over the x-axis. Edmentum answer: The variable k changes as the center of the circle moves vertically. The sign of k is positive when the center is below the x-axis and negative when it is above the x-axis. The sign of the variable k flips when the center moves across the x-axis.

Using the standard form of the equation in question 1, what are the length of the radius and the coordinates of the center for this particular circle? Watch your signs for the variables h and k.

When I compare my equation with the standard form, (x − h)2 + (y − k)2 = r2, I get h = 4, k = -6, and r = 5. The center is at (4, -6), and the length of the radius is 5.

Open GeoGebra. Plot the circle from Question 2 by entering the standard form of the equation in the Input window. (Use ^2 to represent an exponent of 2.) Then plot the same circle using the general form of the equation that was given in Question 1. (Placing your cursor over an equation in the Algebra margin will bold the circle that the equation corresponds to. By right clicking and selecting Equation, you can toggle between the different forms of the equation.) Next, drag the equations of the circles onto the screen from the Algebra margin. Verify that the circles resulting from the two equations are the same. Finally, paste a screenshot of your results in the space below.

c: x^2 + y^2- 8x+ 12y= -27 d: (x-4) ^2+ ( y+6 ) ^2 =25 you have to include a picture

TQ4) What is the general form of the equation of the given circle with center A?

x2 + y2 + 6x - 24y + 128 = 0

The equation of a given circle in general form is x2+ y2 − 8x + 12y + 27 = 0. Write the equation in standard form,(x − h)2 + (y - k)2 = r2, by completing the squares in the equation. Show your work in the table.

x2 + y2 - 8x + 12y + 27 = 0 given x2 - 8x + y2 + 12y = -27 Isolate the constant term x2 - 8x + 16 + y2 + 12y + 36 = -27 + 16 + 36 Complete the square by adding 16 and 36 to both sides. (x2 - 2∙4∙x + 42) + y2 + 12y + 36 = 25 Group and rearrange terms in x. (x - 4)2 + y2 + 12y + 36 = 25 (a - b)2 = a2 - 2ab + b2 (x - 4)2 + (y2 + 2∙6∙y + 62) = 25 Group and rearrange terms in y. (x - 4)2 + (y + 6)2 = 25 (a + b)2 = a2 + 2ab + b2 (x - 4)2 + (y + 6)2 = 52 Take the square root to find r = 5.

TQ3) What is the general form of the equation for the given circle centered at O(0, 0)?

x2 + y2 − 41 = 0

TQ1) What is the general form of the equation of a circle with center at (a, b) and radius of length m?

x^2 + y^2 − 2ax − 2by + (a^2 + b^2 − m^2) = 0