The given equation of the circle is:
[tex]4x^2+4y^2-24x+16y=48[/tex]We can start by simplifying this equation by dividing both sides by 4:
[tex]\begin{gathered} \frac{4x^2}{4}+\frac{4y^2}{4}-\frac{24x}{4}+\frac{16y}{4}=\frac{48}{4} \\ x^2+y^2-6x+4y=12 \end{gathered}[/tex]Now, if we subtract 12 from both sides it is written in general form:
[tex]\begin{gathered} x^2+y^2-6x+4y-12=0 \\ \text{ General form:} \\ x^2+y^2+ax+by+c=0 \end{gathered}[/tex]Where a, b and c correspond to:
[tex]\begin{gathered} a=-2x_0 \\ b=-2y_0 \\ c=x_0^2+y_0^2-r^2 \end{gathered}[/tex]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:
[tex]\begin{gathered} -6=-2x_0 \\ x_0=\frac{-6}{-2} \\ x_0=3 \\ \\ 4=-2y_0 \\ y_0=\frac{4}{-2} \\ y_0=-2 \end{gathered}[/tex]The center of the circle is located at (3,-2).
Let's find the radius:
[tex]\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=\sqrt{25} \\ r=5 \end{gathered}[/tex]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:
[tex]\begin{gathered} 0^2+y^2-6*0+4*y-12=0 \\ y^2+4y-12=0 \\ Apply\text{ the quadratic equation:} \\ y=\frac{-4\pm\sqrt{4^2-4(1)(-12)}}{2(1)} \\ y=\frac{-4\pm\sqrt{64}}{2} \\ y=\frac{-4\pm8}{2} \\ y=\frac{-4-8}{2}=\frac{-12}{2}=-6\text{ and }y=\frac{-4+8}{2}=\frac{4}{2}=2 \end{gathered}[/tex]The y-intercepts are y=-6 and y=2.
Now let's find the x-intercepts:
[tex]\begin{gathered} x^2+0^2-6x+4*0-12=0 \\ x^2-6x-12=0 \\ \text{ Apply the quadratic equation:} \\ x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(-12)}}{2(1)} \\ x=\frac{6\pm\sqrt{84}}{2} \\ x=\frac{6\pm9.2}{2} \\ x=\frac{6-9.2}{2}=\frac{-3.2}{2}=-1.6\text{ and }x=\frac{6+9.2}{2}=\frac{15.2}{2}=7.6 \end{gathered}[/tex]The x-intercepts are x=-1.6 and x=7.6
