Q 30. - Precalculus Enhanced with Graphing Utilities

Question

In Problem, (a) find the center $\left(h,k\right)$ and radius $r$ of each circle; (b) graph each circle; (c) find the intercepts, if any.

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

Step-by-Step Solution

Answer

(a) Center is $\left(-\frac{1}{2},-\frac{1}{2}\right)$ and radius is $1$ unit.

(b) Graph is as follows:

(c) $x-$intercepts are $\left(\frac{-1\pm \sqrt{3}}{2},0\right)$ and the $y-$intercepts are $\left(0,\frac{-1\pm \sqrt{3}}{2}\right)$

Step 1. Given information

An equation of a circle is $x^2+y^2+x+y-\frac{1}{2}=0$.

Step 2. Center and radius of a circle.

Consider the equation of a circle: $x^2+y^2+x+y-\frac{1}{2}=0$

$\begin{array}{rcl}\left(x^2+2·\frac{1}{2}·x+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2\right)+\left(y^2+2·\frac{1}{2}·y+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2\right)-\frac{1}{2}& =& 0\\ \left(x^2+2·\frac{1}{2}·x+{\left(\frac{1}{2}\right)}^2\right)-{\left(\frac{1}{2}\right)}^2+\left(y^2+2·\frac{1}{2}·y+{\left(\frac{1}{2}\right)}^2\right)-{\left(\frac{1}{2}\right)}^2-\frac{1}{2}& =& 0\\ {\left(x+\frac{1}{2}\right)}^2+{\left(y+\frac{1}{2}\right)}^2-1& =& 0\\ {\left(x+\frac{1}{2}\right)}^2+{\left(y+\frac{1}{2}\right)}^2& =& 1^2\end{array}$

The above equation is standard form of the circle with radius $1$ unit and center $\left(-\frac{1}{2},-\frac{1}{2}\right)$.

Step 3. Graph of the circle

Graph of the circle is as follows:

Step 4. Intercepts of the circle.

To find the $x-$intercepts, substitute $y=0$ and solve for .

$\begin{array}{rcl}{\left(x+\frac{1}{2}\right)}^2+{\left(0+\frac{1}{2}\right)}^2& =& 1^2\\ {\left(x+\frac{1}{2}\right)}^2+\frac{1}{4}& =& 1\\ x+\frac{1}{2}& =& \pm \frac{\sqrt{3}}{2}\\ x& =& \frac{-1\pm \sqrt{3}}{2}\end{array}$

Therefore, the $x-$intercepts are $\left(\frac{-1\pm \sqrt{3}}{2},0\right)$.

To find the $y-$intercepts, substitute $x=0$ and solve for $y$.

$\begin{array}{rcl}{\left(0+\frac{1}{2}\right)}^2+{\left(y+\frac{1}{2}\right)}^2& =& 1^2\\ y+\frac{1}{2}& =& \pm \frac{\sqrt{3}}{2}\\ y& =& \frac{-1\pm \sqrt{3}}{2}\end{array}$

Therefore, the $y-$intercepts are $\left(0,\frac{-1\pm \sqrt{3}}{2}\right)$.