Chapter 1.8, Problem 108E

To determine

To show: The equation $3x^2+3y^2+6x-y=0$ represents a circle and evaluate the center and radius of the circle.

Answer to Problem 108E

The equation $3x^2+3y^2+6x-y=0$ represents a circle. The center of the circle is $\left(-1,\frac{1}{6}\right)$ and radius is $\frac{\sqrt{37}}{36}$ units.

Explanation of Solution

Given information:

The equation $3x^2+3y^2+6x-y=0$ .

Formula used:

In order to solve a quadratic equation, completing the square method is used that transforms the equation in the form of square trinomial.

Step1: Divide the equation by coefficient of $x^2$ if it is not equal to 1.

Step 2: Take the square of the half of the coefficient of x and add it to both sides of the equation.

Step 3: Factor the equation.

The standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Calculation:

Consider the equation $3x^2+3y^2+6x-y=0$ .

Recall that in order to solve a quadratic equation, completing the square method is used that transforms the equation in the form of square trinomial.

Step1: Divide the equation by coefficient of $x^2$ if it is not equal to 1.

Step 2: Take the square of the half of the coefficient of x and add it to both sides of the equation.

Step 3: Factor the equation.

In the provided equation, divide the entire equation by 3,

  $x^2+y^2+2x-\frac{y}{3}=0$

Add 1 to both the sides of the equation,

  $x^2+y^2+2x-\frac{y}{3}+1=0+1$

Again add $\frac{1}{36}$ to both the sides of the equation,

  $x^2+y^2+2x-\frac{y}{3}+1+\frac{1}{36}=0+1+\frac{1}{36}$

Group the terms,

  $\begin{array}{c}{x}^2+y^2+2x-\frac{y}{3}+1+\frac{1}{36}=0+1+\frac{1}{36}\\ \left(x^2+2x+1\right)+\left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{37}{36}\end{array}$

Factor out the trinomial, recall that ${\left(a+b\right)}^2=a^2+2ab+b^2$ and ${\left(a-b\right)}^2=a^2-2ab+b^2$ .

Apply it,

  $\begin{array}{c}{x}^2+y^2+2x-\frac{y}{3}+1+\frac{1}{36}=0+1+\frac{1}{36}\\ \left(x^2+2x+1\right)+\left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{37}{36}\\ {\left(x+1\right)}^2+{\left(y-\frac{1}{6}\right)}^2=\frac{37}{36}\end{array}$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Convert the equation obtained above in standard form,

  $\begin{array}{c}{x}^2+y^2+2x-\frac{y}{3}+1+\frac{1}{36}=0+1+\frac{1}{36}\\ \left(x^2+2x+1\right)+\left(y^2-\frac{y}{3}+\frac{1}{36}\right)=\frac{37}{36}\\ {\left(x+1\right)}^2+{\left(y-\frac{1}{6}\right)}^2=\frac{37}{36}\\ {\left(x-\left(-1\right)\right)}^2+{\left(y-\frac{1}{6}\right)}^2={\left(\frac{\sqrt{37}}{6}\right)}^2\end{array}$

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-\left(-1\right)\right)}^2+{\left(y-\frac{1}{6}\right)}^2={\left(\frac{\sqrt{37}}{6}\right)}^2$ .

Here, $h=-1,k=\frac{1}{6}\text{ and }r=\frac{\sqrt{37}}{36}$ .

Therefore, center of circle is $\left(-1,\frac{1}{6}\right)$ and radius is $\frac{\sqrt{37}}{36}$ .

Thus, the equation $3x^2+3y^2+6x-y=0$ represents a circle. The center of the circle is $\left(-1,\frac{1}{6}\right)$ and radius is $\frac{\sqrt{37}}{36}$ units.