Question

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

Answer 1

Answer:

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

Step-by-step explanation:

Because the equation of the circle is found by using the center and radius only.The equation of any circle with center (h,k) and of radius r is (x-h)^2+(y-k)^2=r^2.

Answer 2

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

$\displaystyle \sf{ {x}^{2} + {y}^{2} + 4x - 8y + 11 = 0 }$

$\displaystyle \sf{ \implies {x}^{2} + 4x + {y}^{2} - 8y + 11 = 0}$

$\displaystyle \sf{ \implies {x}^{2} + 2.x.2 + {2}^{2} + {y}^{2} - 2.y .4 + {4}^{2} - {2}^{2} - {4}^{2} + 11 = 0}$

$\displaystyle \sf{ \implies {(x + 2)}^{2} + {(y - 4)}^{2} -4 - 16 + 11 = 0}$

$\displaystyle \sf{ \implies {(x + 2)}^{2} + {(y - 4)}^{2} - 9= 0}$

$\displaystyle \sf{ \implies {(x + 2)}^{2} + {(y - 4)}^{2} = 9}$

$\displaystyle \sf{ \implies {(x + 2)}^{2} + {(y - 4)}^{2} = {3}^{2} }$

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)

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