Q 25. - 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.

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

Step-by-Step Solution

Answer

(a) Center of a circle is $\left(1,2\right)$ and radius is$3$ units.

(b) Graph of a circle is as follows:

(c) $x-$intercepts are $\left(1\pm \sqrt{5},0\right)$ and $y-$intercepts are $\left(0,2\pm 2\sqrt{2}\right)$

Step 1. Given information

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

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 $(x-h{)}^2+(y-k{)}^2=r^2$.

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

The above equation is standard form of the circle with radius $3$ and center $(1,2)$.

Step 3. Graph of a circle

Graph is as follows:

Step 4. Find the intercepts.

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

Therefore, the $x-$intercepts are $(1\pm \sqrt{5},0)$

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

$\begin{array}{rcl}(0-1{)}^2+(y-2{)}^2& =& 3^2\\ 1+(y-2{)}^2& =& 9\\ y-2& =& \pm \sqrt{8}\\ y-2& =& \pm 2\sqrt{2}\\ y& =& 2\pm 2\sqrt{2}\end{array}$

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