Chapter 1.8, Problem 106E

To determine

To show: The equation $x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}=0$ represents a circle and evaluate the center and radius of the circle.

Answer to Problem 106E

The equation $x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}=0$ represents a circle. The center of the circle is $\left(-\frac{1}{4},-1\right)$ and radius is $1$ units.

Explanation of Solution

Given information:

The equation $x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}=0$ .

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 $x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}=0$ .

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, add $\frac{1}{16}$ to both the sides of the equation,

  $x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}+\frac{1}{16}=\frac{1}{16}$

Again add 1 to both the sides of the equation,

  $x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}+\frac{1}{16}+1=\frac{1}{16}+1$

Group the terms,

  $\begin{array}{c}{x}^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}+\frac{1}{16}+1=\frac{1}{16}+1\\ \left(x^2+\frac{1}{2}x+\frac{1}{16}\right)+\left(y^2+2y+1\right)=1\end{array}$

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,

  $\begin{array}{c}{x}^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}+\frac{1}{16}+1=\frac{1}{16}+1\\ \left(x^2+\frac{1}{2}x+\frac{1}{16}\right)+\left(y^2+2y+1\right)=1\\ {\left(x+\frac{1}{4}\right)}^2+{\left(y+1\right)}^2=1\end{array}$

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,

  $\begin{array}{c}{x}^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}+\frac{1}{16}+1=\frac{1}{16}+1\\ \left(x^2+\frac{1}{2}x+\frac{1}{16}\right)+\left(y^2+2y+1\right)=1\\ {\left(x+\frac{1}{4}\right)}^2+{\left(y+1\right)}^2=1\\ {\left(x-\left(-\frac{1}{4}\right)\right)}^2+{\left(y-\left(-1\right)\right)}^2={\left(1\right)}^2\end{array}$

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-\left(-\frac{1}{4}\right)\right)}^2+{\left(y-\left(-1\right)\right)}^2={\left(1\right)}^2$ .

Here, $h=-\frac{1}{4},k=-1\text{ and }r=1$ .

Therefore, center of circle is $\left(-\frac{1}{4},-1\right)$ and radius is $1$ .

Thus, the equation $x^2+y^2+\frac{1}{2}x+2y+\frac{1}{16}=0$ represents a circle. The center of the circle is $\left(-\frac{1}{4},-1\right)$ and radius is $1$ units.