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Chapter 6: Problem 74

Will help you prepare for the material covered in the next section. Complete the square and write the equation in standard form: \(x^{2}+6 x+y^{2}=0 .\) Then give the center and radius of the circle, and graph the equation.

Short Answer

Expert verified
The equation in standard form is \((x+3)^{2} + y^{2} = 9\), the center of the circle is \((-3,0)\), and the radius is 3.

Step by step solution

01

Completing the square for x term

To complete the square, half of the coefficient of \(x\) is squared and added to both sides of the equation. In this case, the coefficient of \(x\) is 6, so \((6/2)^{2}=9\). Then the equation becomes \(x^{2}+6 x+9+y^{2}=-9+9\)
02

Rewrite the equation in standard form

The equation can be rewritten as \((x+3)^{2} + y^{2} = 9\) by recognizing that \((x+3)^{2}\) is the square of \(x+3\). This is the standard form equation of a circle, \((x-h)^{2} + (y-k)^{2} = r^{2}\), where \((h,k)\) are the coordinates of the center and \(r\) is the radius of the circle.
03

Identify the center and radius

Comparing the standard form equation with our new equation, it can be seen that \(-3\) is the x-coordinate of the center while the y-coordinate is 0 because \(y\) is not accompanied by a constant. So, the center of the circle is \((-3,0)\). The radius of the circle is the square root of 9, which is 3. So the radius is \(r=3\).
04

Graph the equation

First plot the center of the circle at point \((-3,0)\). Then, draw a circle with radius 3 units, centered at \((-3,0)\).

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Most popular questions from this chapter

How do you determine if two vectors are orthogonal?

Find \(\text {pro}_{\mathbf{w}} \mathbf{V}\) Then decompose v into two vectors, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2},\) where \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\) and \(\mathbf{v}_{2}\) is orthogonal to \(\mathbf{w}.\) $$\mathbf{v}=2 \mathbf{i}+\mathbf{j}, \quad \mathbf{w}=6 \mathbf{i}+3 \mathbf{j}$$

Find the angle between \(\mathbf{v}\) and \(\mathbf{w} .\) Round to the nearest tenth of a degree. $$\mathbf{v}=-3 \mathbf{i}+2 \mathbf{j}, \quad \mathbf{w}=4 \mathbf{i}-\mathbf{j}$$

Use the vectors $$\mathbf{u}=a_{1} \mathbf{i}+b_{1} \mathbf{j}, \quad \mathbf{v}=a_{2} \mathbf{i}+b_{2} \mathbf{j}, \quad \text { and } \quad \mathbf{w}=a_{3} \mathbf{i}+b_{3} \mathbf{j},$$ to prove the given property. $$\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}$$

How do you determine the work done by a force F in moving an object from \(A\) to \(B\) when the direction of the force is not along the line of motion?
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