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Find the center and radius of a circle modeled by the equation x2 + y2 - 4x + 2y - 4 = 0 by completing the square

User Song Bee
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3 votes
ANSWER

Center: (2,-1)

Radius: 3 units.

Step-by-step explanation

The given circle has equation:


{x}^(2) + {y}^(2) - 4x + 2y - 4 = 0

We rearrange to get;


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

We add the square of half the coefficients of the linear terms to both sides of the equation as shown below:


{x}^(2) - 4x + {( - 2)}^(2) + {y}^(2)+ 2y + {(1)}^(2) =4+ {( - 2)}^(2)+ {(1)}^(2)

Look out for the perfect squares on the left hand side .


{(x - 2)}^(2) +{(y + 1)}^(2) =4+ 4+ 1


{(x - 2)}^(2) +{(y + 1)}^(2) =9


{(x - 2)}^(2) +{(y + 1)}^(2) = {3}^(2)

By comparing to :


{(x - h)}^(2) +{(y - k)}^(2) = {r}^(2)

We have the center to be

(h,k)=(2,-1) and radius r=3.
User Barfatchen
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