Question

for the 4x^2+4y^2-24X+16Y=48 A. Find the center and radius r of the circleB Graph the circleC. find the intercepts

Answer 1

The given equation of the circle is:

`4x^2+4y^2-24x+16y=48`

We can start by simplifying this equation by dividing both sides by 4:

`\begin{gathered} (4x^2)/(4)+(4y^2)/(4)-(24x)/(4)+(16y)/(4)=(48)/(4) \\ x^2+y^2-6x+4y=12 \end{gathered}`

Now, if we subtract 12 from both sides it is written in general form:

`\begin{gathered} x^2+y^2-6x+4y-12=0 \\ \text{ General form:} \\ x^2+y^2+ax+by+c=0 \end{gathered}`

Where a, b and c correspond to:

`\begin{gathered} a=-2x_0 \\ b=-2y_0 \\ c=x_0^2+y_0^2-r^2 \end{gathered}`

The coordinates of the center of the circle are (x0,y0) and r is the radius.

By replacing the known values we can obtain the center and radius as follows:

`\begin{gathered} -6=-2x_0 \\ x_0=(-6)/(-2) \\ x_0=3 \\ \\ 4=-2y_0 \\ y_0=(4)/(-2) \\ y_0=-2 \end{gathered}`

The center of the circle is located at (3,-2).

Let's find the radius:

`\begin{gathered} -12=3^2+(-2)^2-r^2 \\ -12=9+4-r^2 \\ -12=13-r^2 \\ -12-13=-r^2 \\ -25=-r^2 \\ r^2=25 \\ r=√(25) \\ r=5 \end{gathered}`

The radius of the circle is 5.

PART B.

By knowing the coordinates of the center and the radius, the graph of the circle looks like this:

PART C.

To find the intercepts, we need to make x=0 and solve for y, and then make y=0 and solve for x as follows:

`\begin{gathered} 0^2+y^2-6*0+4*y-12=0 \\ y^2+4y-12=0 \\ Apply\text{ the quadratic equation:} \\ y=(-4\pm√(4^2-4(1)(-12)))/(2(1)) \\ y=(-4\pm√(64))/(2) \\ y=(-4\pm8)/(2) \\ y=(-4-8)/(2)=(-12)/(2)=-6\text{ and }y=(-4+8)/(2)=(4)/(2)=2 \end{gathered}`

The y-intercepts are y=-6 and y=2.

Now let's find the x-intercepts:

`\begin{gathered} x^2+0^2-6x+4*0-12=0 \\ x^2-6x-12=0 \\ \text{ Apply the quadratic equation:} \\ x=(-(-6)\pm√((-6)^2-4(1)(-12)))/(2(1)) \\ x=(6\pm√(84))/(2) \\ x=(6\pm9.2)/(2) \\ x=(6-9.2)/(2)=(-3.2)/(2)=-1.6\text{ and }x=(6+9.2)/(2)=(15.2)/(2)=7.6 \end{gathered}`

The x-intercepts are x=-1.6 and x=7.6