Q 27. - Precalculus Enhanced with Graphing Utilities

Question

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

$x^2+y^2+4x-4y-1=0$

Step-by-Step Solution

Answer

(a) Center is $\left(-2,2\right)$and radius is equal to $3$ units.

(b) Graph is as follows:

(c) $x-$intercepts are $(-2\pm \sqrt{5},0)$ and $y-$intercepts are $(0,2\pm \sqrt{5})$

Step 1. Given information

An equation of a circle is $x^2+y^2+4x-4y-1=0$.

Step 2. Center of the circle

Consider an equation of a circle: $x^2+y^2+4x-4y-1=0$

$\begin{array}{rcl}(x+2{)}^2+(y-2{)}^2& =& 9\\ (x+2{)}^2+(y-2{)}^2& =& 3^2\end{array}$

The above equation is standard form of the circle with radius $3$ and center $\left(-2,2\right)$.

Step 3. Graph of the circle.

Graph is as follows:

Step 4. Intercepts of the circle

Consider the equation: $(x+2{)}^2+(y-2{)}^2=3^2$

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

$\begin{array}{rcl}(x+2{)}^2+(0-2{)}^2& =& 3^2\\ (x+2{)}^2+4& =& 9\\ x+2& =& \pm \sqrt{5}\\ x& =& -2\pm \sqrt{5}\end{array}$

Therefore, the $x-$intercepts are $(-2\pm \sqrt{5},0)$

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

$\begin{array}{rcl}(0+2{)}^2+(y-2{)}^2& =& 3^2\\ 4+(y-2{)}^2& =& 9\\ y-2& =& \pm \sqrt{5}\\ y& =& 2\pm \sqrt{5}\end{array}$

Therefore, the $y-$intercepts are $(0,2\pm \sqrt{5})$