Q 24. - 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.

$3(x+1{)}^2+3(y-1{)}^2=6$

Step-by-Step Solution

Answer

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

(b) Graph is as follows:

(c) $x-$intercepts are $-2,0$ and $y-$intercepts are $0,2$

Step 1. Given information

Equation of a circle is $3(x+1{)}^2+3(y-1{)}^2=6$.

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

The standard form of an equation of a circle with radius $r$ and center $\left(h,k\right)$ is

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

$\begin{array}{rcl}3(x+1{)}^2+3(y-1{)}^2& =& 6\\ (x+1{)}^2+(y-1{)}^2& =& 2\end{array}$

We can see that the center is $(-1,1)$ and the radius is $\sqrt{2}$ units.

Step 3. Graph of the circle

Graph is as follows:

Step 4. Find the intercepts of a circle.

To find $y-$intercepts we let $x=0$ and solve for $y$.

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

Therefore, the $y-$intercepts are $0,2$.

To find $x-$intercepts we let $y=0$ and solve for $x$.

$\begin{array}{rcl}(x+1{)}^2+(0-1{)}^2& =& 2\\ (x+1{)}^2& =& 1\\ x+1& =& \pm 1\\ x& =& -2,0\end{array}$

Therefore, the $y-$intercepts are $0,2$