write x^2+y^2+2x-4y-4=0 in standard form

Question

find the standard form and the center and the radius

Answer 1

Standard form of circle is x^2+y^2+2gx+2fy+c = 0

Given equation x^2+y^2+2x-4y-4 = 0

Compare it to standard form,then

2g = 2, 2f = -4, c = -4

g = 1, f = -2

Center of given circle is (h,k) = (-g,-f ) = (-1,2)

Radius is r = √(g^2+f^2-c)

= √[1^2+(-2)^2-(-4)]

= √(1+4+4)

= √9

= 3

Radius is 3.

Comments

x ² +y ²+ 2gx + 2fy + c = 0 is general equation of a circle.

Standard form of a circle equation is .

Answer 2

The equation is x ² + y ² + 2x - 4y - 4 = 0

x ² + 2x + y ²- 4y  = 4

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here x coefficient = 2. so, (half the x coefficient)² = (2/2)²= 1

Here y coefficient = -4. so, (half the y coefficient)² = (-4/2)²= 4

Add 2 and 1 to each side

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

x ² + 2x + 1 + y ²- 4y + 4 = 9

(x + 1)² + (y - 2)² = 3²

(x - (-1))² + (y - 2)² = 3²

Compare it to standard form of a circle equation is

Where center is (h ,k ) and radius of circle is r .

Radius of circle = 3 and center of the circle is (-1, 2).