Q 29. - 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.

$x^2+y^2-x+2y+1=0$

Step-by-Step Solution

Answer

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

(b) Graph of a circle is as follows:

(c) No $x-$intercept and $y-$intercept is $(0,-1)$

Step 1. Given information

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

Step 2. Center and radius of a circle.

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

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

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

Step 2. Graph of the circle.

Graph is as follows:

Step 4. Intercepts of the circle.

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

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

This equation has no solution because square of a number is never negative. Therefore, there is no $x-$intercept.

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

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

Therefore, the $y-$intercept is $(0,-1)$