Question

Find (a) the center (h, k) and radius r of each circle; (b) graph each circle; (c) find the intercepts if any of

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

Short 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

Step-by-step solution

 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$$\begin{gathered} 2x^2-12x+2y^2+8y=24\sin ce2iscommonfactorsodividedby2 \\ {x}^2-6x+y^2+4y=12,nowwriteequationinsidetheparenthesis \\ (x^2-6x)+\left(y^2+4y\right)=12 \end{gathered}$$

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

complete the square of each expression in parentheses.

$$\begin{gathered} (x^2-6x+9)+(y^2+4y+4)=12+9+4 \\ \sin ceanynumberaddedontheleftsideoftheequationmustbeaddedontheright. \\ {(x-3)}^2+{(y+2)}^2=25,{\left(\frac{-6}{2}\right)}^2=9,{\left(\frac{4}{2}\right)}^2=4 \\ {(x-3)}^2+{(y+2)}^2=5^{2}whichisinthes\tan dardformofcircleequation \\ {(x-h)}^2+{(y-k)}^2=r^2,(h,k)isthecenterandristheradius \end{gathered}$$

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,

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

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 $$\begin{gathered} {(0-3)}^2+{(y+2)}^2=5^2 \\ 9+{(y+2)}^2=25{(y+2)}^2=25-9 \\ {(y+2)}^2=16y+2=\pm 4takesqureroot \\ y=+4-2,-4-2 \\ y=2,-6 \end{gathered}$$

Thus y-intercept points are 2 and -6