# What is the center of a circle whose equation is x^{2} + y^{2} + 4x - 8y + 11 = 0?

**Solution:**

Given equation is x^{2} + y^{2} + 4x - 8y + 11 = 0

Comparing with the general equation of a circle, x^{2} + y^{2} + 2gx + 2fy + c = 0,

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

g = 2, f = -4

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

Aliter

Given equation is x^{2} + y^{2} + 4x - 8y + 11 = 0

x^{2} + 4x + 4 + y^{2} - 8y + 16 = -11 + 4 + 16

(x + 2)^{2 }+ (y - 4)^{2} = 9

(x + 2)^{2} + (y - 4)^{2} = 3^{2}

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

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

## What is the center of a circle whose equation is x^{2} + y^{2} + 4x - 8y + 11 = 0?

**Summary:**

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