WRITE THE FOLLOWING EQUATION IN STANDARD FORM x^2+6x+y^2+4y=0 AND IDENTIFY THE CENTER AND RADIUS

Question

Need to write in Standard Form and Identify the Center and Radius.

.X^2 + 6X +Y^2 + 4Y = 0

Answer 1

Circle equation is x^2+y^2+2gx+2fy+c = 0

x^2+6x+y^2+4y = 0

compare it to above equation.

a = 1, b = 1, c = 0.

2g = 6, 2f = 4

g = 3, f = 2

Center = (-g,-f) = (-3,-2)

r = √(g^2+f^2-c)

r = √(3^2+2^2-0)

r = √(9+4)

radius = √13

Comments

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.

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 + 6x + y² + 4y = 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 = 6, so, (half the x coefficient)² = (6/2)²= 9.

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

Add 9 and 4 to each side.

x² + 6x + 9 + y² + 4y + 4 = 0 + 9 + 4

(x + 3)² + (y + 2)² = 13

(x - (- 3))² + (y - (- 2))² = (√ 13)²

Compare the equation with standard form of a circle equation.

The center is (- 3, - 2) and

The radius is √ 13 units.