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.

$x^2+y^2=4$

Short answer

(a) Center is $\left(0,0\right)$and radius is $2$units

(b) Graph is as follows:

(c) $x-$Intercepts are $\left(\pm 2,0\right)$and $y-$intercepts are$\left(0,\pm 2\right)$

Step-by-step solution

Step 1. Given information

It is given that$x^2+y^2=4$.

Step 2. Find the center and radius.

Consider the equation of a circle.

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

The above equation is standard form of the circle with radius $2$and center$\left(0,0\right)$

Step 3. Graph of the circle

Graph is as follows:

Step 4. Find the intercepts.

Consider the equation $x^2+y^2=4$

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

$\begin{array}{rcl}{x}^2+0^2& =& 4\\ x^2& =& 4\end{array}$

Take square root of both sides

$\begin{array}{rcl}x& =& \pm \sqrt{4}\\ x& =& \pm 2\end{array}$

Therefore, the $x$-intercepts are $(\pm 2,0)$

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

$\begin{array}{rcl}{0}^2+y^2& =& 4\\ y^2& =& 4\end{array}$

Take square root of both sides

$\begin{array}{rcl}y& =& \pm \sqrt{4}\\ y& =& \pm 2\end{array}$

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