Q 23. - 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.

$2(x-3{)}^2+2y^2=8$

Step-by-Step Solution

Answer

(a) Center is $\left(3,0\right)$ and radius is $2$ units.

(b) Graph is as follows:

(c) $x-$intercepts are $\left(1,0\right),\left(5,0\right)$ and there is no $y-$intercept.

Step 1. Given information

It is given that equation of a circle is $2(x-3{)}^2+2y^2=8$.

Step 2. Find the center and the radius.

The standard form the circle with center $\left(h,k\right)$ and radius $r$ is $(x-h{)}^2+(y-k{)}^2=r^2$

$2(x-3{)}^2+2y^2=8$

Divide both sides by $2$.

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

The above equation is standard form of the circle with radius $2$ and center $(3,0)$.

Step 3. Graph of the circle

Graph is as follows:

Step 4. Find the intercepts.

Consider the equation $2(x-3{)}^2+2y^2=8$.

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

$2(x-3{)}^2+2(0{)}^2=8$

Divide both sides by $2$.

$(x-3{)}^2=4$

Take square root of both sides

Add $3$ on both sides

$$\begin{gathered} x=3\pm 2 \\ x=1,5 \end{gathered}$$

Therefore, the $x-$intercepts are $\left(1,0\right)$ and $\left(5,0\right)$

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

$\begin{array}{rcl}2(0-3{)}^2+2y^2& =& 8\\ 18+2y^2& =& 8\end{array}$

Subtract $18$ from both sides

$2y^2=-10$

Divide both sides by $2$

$y^2=-5$

Note that $y^2$ is never negative, because square of a number is either $0$ or a positive number.

Therefore, this equation has no solution and there is no $y-$intercept