Q 26. - 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.

$x^2+y^2+4x+2y-20=0$

Step-by-Step Solution

Answer

(a) Center is $\left(-2,-1\right)$ and radius is equal to $5$ units.

(b) Graph is as follows:

(c) $x-$intercepts are $\left(-2\pm 2\sqrt{6},0\right)$ and $y-$intercepts are $\left(0,-1\pm \sqrt{21}\right)$

Step 1. Given information

Equation of a circle is $x^2+y^2+4x+2y-20=0$.

Step 2. Find the center and radius of a circle.

The standard form the circle with center $\left(h,k\right)$ and radius $r$ is $5$ units.

$\begin{array}{rcl}{x}^2+y^2+4x+2y-20& =& 0\\ \left(x^2+2·2·x+2^2\right)-2^2+\left(y^2+2·1·y+1^2\right)-1^2-20& =& 0\\ (x+2{)}^2-4+(y+1{)}^2-1-20& =& 0\\ (x+2{)}^2+(y+1{)}^2& =& 25\\ (x+2{)}^2+(y+1{)}^2& =& 5^2\\ (x+2{)}^2+(y+1{)}^2& =& 5^2\end{array}$

The above equation is standard form of the circle with radius $5$ and center $(-2,-1)$.

Step 3. Graph of the circle.

Graph is as follows:

Step 4. Find the intercepts of the circle.

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

$\begin{array}{rcl}(x+2{)}^2+(0+1{)}^2& =& 5^2\\ (x+2{)}^2+1& =& 25\\ x+2& =& \pm \sqrt{24}\\ x+2& =& \pm 2\sqrt{6}\\ x& =& -2\pm 2\sqrt{6}\end{array}$

Therefore, the $x-$intercepts are $(-2\pm 2\sqrt{6},0)$

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

$\begin{array}{rcl}(0+2{)}^2+(y+1{)}^2& =& 5^2\\ 4+(y+1{)}^2& =& 25\\ y+1& =& \pm \sqrt{21}\\ y& =& -1\pm \sqrt{21}\end{array}$

Therefore, the $y-$intercepts are $(0,-1\pm \sqrt{21})$