Find the center of the circle x^2 + y^2 +4y -21 =0

Question

I don't know how to find the center of that circle.

Answer 1

Given equation x^2+y^2+4y-21 = 0

x^2+y^2+4y-21 = 0

Add 4 to each side to complete squre.

x^2+y^2+4y+4-21 = 0+4

x^2+(y+2)^2-21 = 4

Add 21 to each side.

x^2+(y+2)^2-21+21 = 4+21

(x-0)^2+(y-(-2))^2 = 25

(x-0)^2+(y-(-2))^2 = (5)^2

Compare it to standard form of circle equation (x-h)^2+(y-k)^2 = r^2

Center is (h,k) = (0,-2)

Radius is 5.

Answer 2

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 + y² + 4y - 21 = 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 = 0, so, (half the x coefficient)² = (0/2)²= 0.

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

Add 0 and 4 to each side.

x² + 0 + y² + 4y + 4 - 21 = 0 + 0 + 4

(x + 0)² + (y + 2)² = 4 + 21 = 25

(x - 0)² + (y - (- 2))² = 5² .

Compare the equation with standard form of a circle equation.

The center of the cicle (h, k) is (0, - 2).