x^2+y^2+2x+8y-8=0

Question

Complete the square and write the equation of the circle in standard form.  Then determine the center and radius o the circle to graph the equation.
x^(2)+y^(2)+2x+8y-8=0

The equation in standard form is?

Answer 1

x^2+y^2+2x+8y-8  = 0

x^2+2x+y^2+8y-8 = 0

To complete the squre add (half of x coefficient)^2 & (half of y coefficient)^2 to each side.

x^2+2x+1+y^2+8y+16-8 = 1+16

(x+1)^2+(y+4)^2+8 = 17

Subtract 8 from each side.

(x+1)^2+(y+4)^2 = 11

(x-(-1))^2+(y-(-4))^2 = 11

Compare itto circle equation (x-h)^2+(y-k)^2 = r^2

Center is (-1,-4) and radius is r = √11.

Graph

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 + 2x + 8y - 8 = 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 = 2, so, (half the x coefficient)² = (2/2)²= 1.

Here y coefficient = 8, so, (half the y coefficient)² = (8/2)²= 16.

Add 1 and 16 to each side.

x² + 2x + 1 + y² + 8y + 16 - 8 = 0 + 1 + 16

(x + 1)² + (y + 4)² = 17 + 8 = 25

(x - (- 1))² + (y - (- 4))² = 5².

Compare the equation with standard form of a circle equation.

The center (h, k) is (- 1, - 4), and the radius (r) is 5 units.

GRAPH :

1. Draw the coordinate plane.

2. Place the center of the circle at (- 1, - 4).

3. Plot the radius points on the coordinate plane.

   Since the radius is 5 units,

  Count 5 units up, down, left, and right from the center (- 1, - 4).

  This means that we should have the points at (- 1, 1), (- 1, - 9), (- 6, - 4), and (4, - 4).

4. Connect the plotted points to the graph of the circle with a round, smooth  curve.