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Chapter 2: Problem 53

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+6 x+2 y+6=0$$

Short Answer

Expert verified
The standard form of the equation is \((x+3)^2 + (y+1)^2 = 4\). The center of the circle is \((-3, -1)\) and the radius is 2.

Step by step solution

01

Rearrange the Equation

First, rewrite the equation by grouping x terms together, y terms together, and moving constants to the right side of the equation. The equation becomes: \(x^2 + 6x + y^2 + 2y = -6\)
02

Complete the Square for x and y

Use the 'completing the square' method for x and y separately. For x, Halve the coefficient of x which is 6, then square it. \( (6/2)^2 = 9 \). Add 9 on both sides of the equation. Do the same for y, halve the coefficient of y (2), then square it. \( (2/2)^2 = 1 \). Add 1 on both sides of the equation. The equation becomes: \(x^2 + 6x +9 + y^2 + 2y +1 = -6 + 9 + 1\)
03

Rewrite the Equation in Standard Form

Now, rewrite what you have just done in the form \((x-h)^2+ (y-k)^2 = r^2\). So \( (x+3)^2 + (y+1)^2 = 4\).
04

Identify the Center and Radius of the Circle

From the standard form, the center (h, k) of the circle is \((-3, -1)\) and the radius r is \(鈭�4 = 2\).
05

Plot the Circle on a Graph

On a graph, mark the center at point (-3, -1) and draw a circle with radius 2 units around it.
06

Check the Equation

Finally, plugin several points on the circle into your equation. If at least two points satisfy the equation, then your circle is correct.

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