Respuesta :
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).
Answer:
10. (x + 5)² + (y - 3)² = 25
11. (x - 8)² + (y - 15)² = 9
12. (x + 6)² + (y - 11)² = 20
Step-by-step 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