Question

Part 4: Use the information provided to write the standard form equation of each equation

10. (x + 5)^2 + y^2 = 25
Translated 3 up

11. x^2 + y^2 - 16x - 30y + 280 = 0

12. x^2 + y^2 + 12x - 22y + 137 = 0

Answer 1

10. (x + 5)² + (y - 3)² = 25

11. (x - 8)² + (y - 15)² = 9

12. (x + 6)² + (y - 11)² = 20

Explanation:

10. (x + 5)^2 + y^2 = 25

Centre: (-5,0)

Translated 3 up

Centre: (-5,3)

(x + 5)² + (y - 3)² = 25

11. x^2 + y^2 - 16x - 30y + 280 = 0

h = -16/-2 = 8

k = -30/-2 = 15

8² + 15² - 280 = r²

r² = 9

(x - 8)² + (y - 15)² = 9

12. x^2 + y^2 + 12x - 22y + 137 = 0

h = 12/-2 = -6

k = -22/-2 = 11

(-6)² + 11² - 137 = r²

r² = 20

(x - (-6))² + (y - 11)² = 20

(x + 6)² + (y - 11)² = 20

Answer 2

Problem 10

Answer: (x+5)^2+(y-3)^2=25

------------

Work Shown:

All we're doing really is replacing the original 'y' with 'y-3' to shift the center up 3 units. This shifts every point on the circle the same as well.

This new circle has center (-5,3) and radius 5. The old center was (-5,0). The radius stayed the same.

===========================================================

Problem 11

Answer: (x-8)^2+(y-15)^2 = 9

------------

Work Shown:

x^2+y^2-16x-30y+280 = 0

(x^2-16x)+(y^2-30y)+280 = 0

(x^2-16x+64)-64+(y^2-30y+225)-225+280 = 0

(x-8)^2+(y-15)^2-9 = 0

(x-8)^2+(y-15)^2 = 9

This circle has center (8,15) and radius 3.

===========================================================

Problem 12

Answer: (x+6)^2+(y-11)^2 = 20

------------

Work Shown:

x^2+y^2+12x-22y+137 = 0

(x^2+12x)+(y^2-22y)+137 = 0

(x^2+12x+36)-36+(y^2-22y+121)-121+137 = 0

(x+6)^2+(y-11)^2-20 = 0

(x+6)^2+(y-11)^2 = 20

This circle has center (-6,11) and radius sqrt(20) = 2*sqrt(5).