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Find the center of a circle with the equation: x^2+y^2−4x+2y−11=0 A. (1, -2)B. (-1, 2)C. (2, -1)D. (1, 2)

User Umefarooq
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1 Answer

16 votes
16 votes

x^2+y^2−4x+2y−11=0

First, complete squares

x^2-4x +y^2+2y = 11

Determine c1 to complete the first square(x):

c1= (b/2)^2

c1= (-4/2)^2 = (-2)^2 = 4

Then c2 for y

c2= (2/2)^2 = 1^2 = 1

Rewrite:

x^2 -4x +4 +y^2 +2y + 1 = 11 + 4 + 1

Factorize:

(x-2)^2 + (y+1)^2 = 16

Circle equation:

(x-a)^2 + (y-b)^2 = r^2

Where r is the radius

and (a,b) is the center

So, the center of the circle is:

(2,-1) option C

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