Q. 32 - Precalculus Enhanced with Graphing Utilities

Question

Find (a)  the center (h, k) and radius r of each circle; (b) graph each circle; (c) find the intercepts if any of the equation $2x^2+2y^2+8x+7=0$

Step-by-Step Solution

Answer

(a) center (h ,k), and the radius of the given equation  are (-2,0) and$\frac{1}{\sqrt{2}}$

(b) graph of the given equation is 

(c)  x- intercept point is $-2\pm \frac{1}{\sqrt{2}}$

Part(a) Step 1. Given information

Given the equation of the circle is $2x^2+2y^2+8x+7=0$

 We have to   Find (a)  the center (h, k) and radius r of each circle.

Part (a) Step 2. Rewriting the equation of a circle

 Given equation group the expression containing x, group the expression containing y, and put the constant on the other side of the equation. The equation becomes $$\begin{gathered} 2x^2+2y^2+8x+7=0 \\ 2x^2+8x+2y^2=-7 when divide by 2 to write equation in s\tan dard form \\ {x}^2+4x+y^2=\raisebox{1ex}{$-7$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ (x^2+4x)+\left(y^2\right)=\raisebox{1ex}{$-7$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$

Part(a) Step 3. Find the center and radius

complete the square of each expression in parentheses. 

$$\begin{gathered} (x^2+4x)+\left(y^2\right)=\raisebox{1ex}{$-7$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ (x^2+4x+4)+y^2=\raisebox{1ex}{$-7$}\!\left/ \!\raisebox{-1ex}{$2+4$}\right. \\ \sin ce any number added on the left side of the equation must be added on the right. \\ {(x--2)}^2+{(y-0)}^2={\left(\frac{1}{\sqrt{2}}\right)}^2 this is the s\tan datrd form of the equation with center(-2,0) and radius \frac{1}{\sqrt{2}}. \\ So center(-2,0) and radius \frac{1}{\sqrt{2}}. \end{gathered}$$

Part (b) Step 1. Given information

Given equation of a circle  is $2x^2+2y^2+8x+7=0$

 we need to  sketch the graph of a circle

Part (b) Step 2 .Plotting the graph

Given equation  $2x^2+2y^2+8x+7=0$in the form ${(x--2)}^2+{(y-0)}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$

      hence radius $is \frac{1}{\sqrt{2}} and center is (-2,0)$. thus the graph is given below

Part (c) Step 1. Given information

Here equation of a circle  is $2x^2+2y^2+8x+7=0$

 we have to find out the intercept if there is any

Part(c) Step 2. Find the intercept

Given equation in the form 

${(x--2)}^2+{(y-0)}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$

To find x-intercept put y=0 we get

$$\begin{gathered} {(x--2)}^2+{(o-0)}^2={\left(\frac{1}{\sqrt{2}}\right)}^2 \\ {(x+2)}^2=\frac{1}{2} (x+2)=\pm \frac{1}{\sqrt{2}} \\ x=-2\pm \frac{1}{\sqrt{2}},this x- intercept form \end{gathered}$$