Q.31 - 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

$2x^2+2y^2-12x+8y-24=0$

Step-by-Step Solution

Answer

(a) The center of the given equation of the circle is (3,-2) and the radius is 5

(b) graph of a circle is given below

(c)  X intercept point is $3\pm \sqrt{21}$

     y-intercept points are-2  and 6

Part (a) Step 1. Given information

Here given equation of the circle is $2x^2+2y^2-12x+8y-24=0$

We have to find  Find

(a)  the center (h, k)   and radius

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

In the given equation group the expression containing x, group the expression containing y, and put the

constant on the right side of the equation. we get$2x^2-12x+2y^2+8y=24$

Here 2 is a common factor so we can divide by 2 , then the equation becomes$$\begin{gathered} 2x^2-12x+2y^2+8y=24 \sin ce 2 is common factor so divided by 2 \\ {x}^2-6x+y^2+4y=12, now write equation inside the parenthesis \\ (x^2-6x)+\left(y^2+4y\right)=12 \end{gathered}$$

 

Part (a) step 3. (a) Finding center and radius

 complete the square of each expression in parentheses. 

$$\begin{gathered} (x^2-6x+9)+(y^2+4y+4)=12+9+4 \\ \sin ce any number added on the left side of the equation must be added on the right. \\ {(x-3)}^2+{(y+2)}^2=25 ,{\left(\frac{-6}{2}\right)}^2=9,{\left(\frac{4}{2}\right)}^2=4 \\ {(x-3)}^2+{(y+2)}^2=5^2 which is inthe s\tan dard form of circle equation \\ {(x-h)}^2+{(y-k)}^2=r^2,(h, k) is the center and r is the radius \end{gathered}$$

hence the center is(3,-2) and radius is 5 

Part (b) Step 1. Given information

Here given equation of the circle is $2x^2+2y^2-12x+8y-24=0$

We have to sketch the graph of the  circle

Part (b) Step 2 .Plotting the graph

According to the given equation center is (3,-2) and radius is 5

so the graph of the circle with center (3,-2) and radius 5 is given below

Part (c) Step.1 Given information

Here equation is in the form $2x^2+2y^2-12x+8y-24=0$

 we have to find out the intercepts if there is any

Part (c) step 2. Finding X -intercept

The given equation is in the form ${(x-3)}^2+{(y+2)}^2=5^2$.

To find x-intercept put y=0 we get,

$$\begin{gathered} {(x-3)}^2+{(0+2)}^2=5^2 {(x-3)}^2+4=25 \\ {(x-3)}^2=25-4 \\ {(x-3)}^2=21 (x-3=\pm \sqrt{21} \\ x=3\pm \sqrt{21} , this is the x intercept point \end{gathered}$$

Part (c) Step 3. Find the y-intercept

 Here given equation  is in the form${(x-3)}^2+{(y+2)}^2=5^2$

 To get  y-intercept put x=0 then equation becomes  $$\begin{gathered} {(0-3)}^2+{(y+2)}^2=5^2 \\ 9+{(y+2)}^2=25 {(y+2)}^2=25-9 \\ {(y+2)}^2=16 y+2=\pm 4 take squre root \\ y=+4-2 ,-4-2 \\ y=2,-6 \end{gathered}$$

Thus y-intercept points are 2 and -6