MATH SOLVE

3 months ago

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

Accepted Solution

A:

ANSWER

Center: (2,-1)

Radius: 3 units.

EXPLANATION

The given circle has equation:

[tex] {x}^{2} + {y}^{2} - 4x + 2y - 4 = 0[/tex]

We rearrange to get;

[tex]{x}^{2} - 4x + {y}^{2}+ 2y =4[/tex]

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

[tex]{x}^{2} - 4x + {( - 2)}^{2} + {y}^{2}+ 2y + {(1)}^{2} =4+ {( - 2)}^{2}+ {(1)}^{2} [/tex]

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

[tex] {(x - 2)}^{2} +{(y + 1)}^{2} =4+ 4+ 1[/tex]

[tex]{(x - 2)}^{2} +{(y + 1)}^{2} =9[/tex]

[tex]{(x - 2)}^{2} +{(y + 1)}^{2} = {3}^{2} [/tex]

By comparing to :

[tex]{(x - h)}^{2} +{(y - k)}^{2} = {r}^{2} [/tex]

We have the center to be

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

Center: (2,-1)

Radius: 3 units.

EXPLANATION

The given circle has equation:

[tex] {x}^{2} + {y}^{2} - 4x + 2y - 4 = 0[/tex]

We rearrange to get;

[tex]{x}^{2} - 4x + {y}^{2}+ 2y =4[/tex]

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

[tex]{x}^{2} - 4x + {( - 2)}^{2} + {y}^{2}+ 2y + {(1)}^{2} =4+ {( - 2)}^{2}+ {(1)}^{2} [/tex]

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

[tex] {(x - 2)}^{2} +{(y + 1)}^{2} =4+ 4+ 1[/tex]

[tex]{(x - 2)}^{2} +{(y + 1)}^{2} =9[/tex]

[tex]{(x - 2)}^{2} +{(y + 1)}^{2} = {3}^{2} [/tex]

By comparing to :

[tex]{(x - h)}^{2} +{(y - k)}^{2} = {r}^{2} [/tex]

We have the center to be

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