Question: I have the following equation of a circle:

${x^2 + y^2 + 8x + 12y -48 = 0}$

In order to compete the square of x and y I do the following:

${(x + 4)^2 + 16 + (y + 6)^2 + 36 - 48 = + 16 + 36}$

=> ${(x + 4)^2 + (y + 6)^2 + 4 = 52}$

=> ${(x + 4)^2 + (y + 6)^2 = 48}$

I would say the centre is (-4, -6) with a radiums of ${\sqrt 48}$ but the book gives the radius of 10 and the point the same values. I thought I would have to add 36 and 16 to both sides of the equation so I don't understand where the radius has come from

Answer: 
Given $x^2+y^2+8x+12y-48=0\;,$

Now we can write $x^2+8x+y^2+12y=48\Rightarrow \underbrace{x^2+8x+4^2}+\underbrace{y^2+12y+6^2}=48+4^2+6^2$

so we get $\displaystyle (x+4)^2+(y+6)^2=10^2$

so center is at $(-4,-6)$ and Radius $=10$
A simple way to solve such problem is to remember that, if $C=(\alpha,\beta)$ is the center of a circle, and $r$ is the radius, the equation
$$
(x-\alpha)^2+(y-\beta)^2=r^2
$$
become, with a bit of algebra:
$$
x^2+y^2+ax+by+c=0
$$
with $ a=-2\alpha$, $b=-2\beta$ and $c=\alpha^2+\beta^2-r^2$.

So from your equation we see that $\alpha=-a/2=-4$, $\beta=-b/2=-6$ and $r=\sqrt{\alpha^2+\beta^2-c}=\sqrt{16+36+48}=10$ .

And note that if $\alpha^2+\beta^2-c $ is not positive you have not a (real) circle.