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Chapter 2: Problem 49

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

Step by step solution

01

Rearrange the Terms

The given equation is \(x^{2}+y^{2}+6 x+2 y+6=0\). Rearrange the terms so that all terms involving x are together and all terms involving y are together with the constant term on RHS, we get this form: \(x^{2} + 6x + y^{2} + 2y = -6\).
02

Complete the Square

To complete the square, we need to take half of the coefficient of the x and y terms (which are 6 and 2), square it, then add to both sides. Half of 6 is 3 and half of 2 is 1. So, the equation looks like: \[(x^{2} + 6x + 9) + (y^{2} + 2y + 1) = -6 + 9 + 1\]. This simplifies to: (x+3)^{2} + (y+1)^{2} = 4. Now, the equation is in standard form.
03

Find the Center and Radius

From the standard form equation, the center of the circle is at -a, -b, so the center of our circle is at (-3, -1). The radius of the circle is the square root of the RHS term, the square root of 4 is 2, so the radius of the circle is 2.
04

Graph the Circle

First, put a point at the center of the circle (-3,-1). Then, make another point at 2 units distance from the center in all directions which is the radius of the circle. And finally join these points in a circular path to draw the circle.

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