# what is the center and radius of the circle x^2 + 4x + y^2 + 2y - 9 = 0?

what is the center and radius of the circle x^2 + 4x + y^2 + 2y - 9 = 0?

+1 vote

Given equation x ^2 + 4x  + y ^2+ 2y -9 = 0

x ^2 + y ^2 + 4x  + 2y -9 = 0

Compare it to circle equation x ^2 + y ^2 + 2gx + 2fy + c = 0.

c  = -9

2g  = 4

g  = 2

2f  = 2

f  = 1

Center ( -g , -f ) = ( -2 , -1 )

Radius = √(g ^2 + f ^2 - c)

= √(4 + 1 + 9) = √14

Radius of circle(r ) = 3.74.

Center = ( -2 , -1 )

r  = 3.74.

The standard form of the circle equation is ( x - h )2 + ( y - k )2 = r2, where, (h, k) is the center of the circle, and r is the radius.

The equation is x2 + 4x + y2 + 2y - 9 = 0.

Write the equation in standard form of a circle.

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 = 4, so, (half the x coefficient)² = (4/2)2= 4.

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

Add 4 and 1 to each side.

x2 + 4x + 4 + y2 + 2y + 1 - 9 = 0 + 4 + 1

(x + 2)2 + (y + 1)2 = 9 + 5 = 14

(x - (- 2))2 + (y - (- 1))2 = (√14)2 .

Compare the equation with standard form of a circle equation.

The center (h, k) is (- 2, - 1), and

The radius (r) is √14 units.