Question

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

Answer 1

The center of the circle is ( - 2,4)

Given :

The equation of a circle is x² + y² + 4x - 8y + 11 = 0

To find :

The center of the circle

Formula :

The equation of any circle with center (h,k) and of radius r is

(x - h)² + (y - k)² = r²

Solution :

Step 1 of 2 :

Write down the given equation of the circle

Here the given equation of the circle is

x² + y² + 4x - 8y + 11 = 0

Step 2 of 2 :

Find center of the circle

Which is of the form (x - h)² + (y - k)² = r²

Where h = - 2 , k = 4 , r = 3

Hence center of the circle is ( - 2,4)

Answer 2

centre = (- 2, 4 )

Explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

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

given

x² + y² + 4x - 8y + 11 = 0 ( subtract 11 from both sides )

x² + y² + 4x - 8y = - 11 ( collect x/ y terms )

x² + 4x + y² - 8y = - 11

to obtain the equation in standard form use the method of completing the square.

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

x² + 2(2)x + 4 + y² + 2(- 4)y + 16 = - 11 + 4 + 16

(x + 2)² + (y - 4)² = 9 ← in standard form

with (h, k ) = (- 2, 4 )

centre = (- 2, 4 )