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What is the center of the circle

2x^{2} +2y^{2}-4x+16y=128
A. (1, -2)
B. (1, 4)
C. (1, -4)
D. (-1, -2)
E. (-1, -4

User Edd Morgan
by
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1 Answer

5 votes

Answer:

C

Explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k)are the coordinates of the centre and r is the radius

Given

2x² + 2y² - 4x + 16y = 128 ( divide through by 2 )

x² + y² - 2x + 8y = 64 ( collect x and y terms together )

x² - 2x + y² + 8y = 64

Using the method of completing the square

add ( half the coefficient of the x/ y terms )² to both sides

x² + 2(- 1)x + 1 + y² + 2(4)y + 16 = 64 + 1 + 16

(x - 1)² + (y + 4)² = 81 ← in standard form

with centre = (h, k ) = (1, - 4 ) → C

User Irfan Harun
by
6.7k points