Chapter 1.8, Problem 120E

To determine

To calculate: The conditions on coefficients a, b and c such that equation $x^2+ax+y^2+by+c=0$ represents a circle, a single point and an empty set and evaluate the center and radius of the circle.

Answer to Problem 120E

The equation $x^2+ax+y^2+by=-c$ represents a circle when $\frac{a^2+b^2-4c}{4}>0$ , a point when $\frac{a^2+b^2-4c}{4}=0$ and an empty when $\frac{a^2+b^2-4c}{4}<0$ . The center of the circle is $\left(-\frac{a}{2},-\frac{b}{2}\right)$ and radius is $\sqrt{\frac{a^2+b^2-4c}{4}}$ .

Explanation of Solution

Given information:

The equation $x^2+ax+y^2+by+c=0$ .

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 $x^2+ax+y^2+by+c=0$ .

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.

Isolate the constant term to right hand side of the equation,

  $x^2+ax+y^2+by=-c$

Next, add $\frac{a^2}{4}\text{ and }\frac{b^2}{4}$ to both the sides of the equation,

  $x^2+ax+y^2+by+\frac{a^2}{4}+\frac{b^2}{4}=-c+\frac{a^2}{4}+\frac{b^2}{4}$

Group the terms,

  $\begin{array}{c}{x}^2+ax+y^2+by+\frac{a^2}{4}+\frac{b^2}{4}=-c+\frac{a^2}{4}+\frac{b^2}{4}\\ \left(x^2+ax+\frac{a^2}{4}\right)+\left(y^2+by+\frac{b^2}{4}\right)=\frac{a^2+b^2-4c}{4}\end{array}$

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,

  $\begin{array}{c}{x}^2+ax+y^2+by+\frac{a^2}{4}+\frac{b^2}{4}=-c+\frac{a^2}{4}+\frac{b^2}{4}\\ \left(x^2+ax+\frac{a^2}{4}\right)+\left(y^2+by+\frac{b^2}{4}\right)=\frac{a^2+b^2-4c}{4}\\ {\left(x+\frac{a}{2}\right)}^2{\left(y+\frac{b}{2}\right)}^2=\frac{a^2+b^2-4c}{4}\end{array}$

Now, the above expression represent a circle when the term on the right hand side is strictly greater than 0 that is $\frac{a^2+b^2-4c}{4}>0$ .

Also, the above expression represent a point when the term on the right hand side is equal to 0 that is $\frac{a^2+b^2-4c}{4}=0$ that is sum of squares of two numbers is equal to 0.

And the above expression represent an empty set when the term on the right hand side is strictly less than 0 that is $\frac{a^2+b^2-4c}{4}<0$ that is sum of squares of two numbers is negative which is not possible.

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,

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

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

Here, $h=-\frac{a}{2},k=-\frac{b}{2}\text{ and }r=\sqrt{\frac{a^2+b^2-4c}{4}}$ .

Therefore, center of circle is $\left(-\frac{a}{2},-\frac{b}{2}\right)$ and radius is $\sqrt{\frac{a^2+b^2-4c}{4}}$ .

Thus, the equation $x^2+ax+y^2+by=-c$ represents a circle when $\frac{a^2+b^2-4c}{4}>0$ , a point when $\frac{a^2+b^2-4c}{4}=0$ and an empty when $\frac{a^2+b^2-4c}{4}<0$ . The center of the circle is $\left(-\frac{a}{2},-\frac{b}{2}\right)$ and radius is $\sqrt{\frac{a^2+b^2-4c}{4}}$ .