What is the center of a circle whose equation is x² + y² + 4x - 8y + 11 = 0?

Solution:

Given equation is x² + y² + 4x - 8y + 11 = 0

Comparing with the general equation of a circle, x² + y² + 2gx + 2fy + c = 0,

2g = 4, 2f = -8, c = 11

g = 2, f = -4

Centre = (-g, -f)= (-2, 4)

Aliter

Given equation is x² + y² + 4x - 8y + 11 = 0

By completing the square,

x² + 4x + 4 + y² - 8y + 16 = -11 + 4 + 16

(x + 2)² + (y - 4)² = 9

(x + 2)² + (y - 4)² = 3²

comparing with equation of a circle (x - h)² + (y - k)² = r² centred at (h, k) and radius 'r',

we get centre = (h, k) = (-2, 4)

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What is the center of a circle whose equation is x² + y² + 4x - 8y + 11 = 0?

Summary:

The center of the circle of the given equation is x² + y² + 4x - 8y + 11 = 0 is (-2, 4).