A confusion in the Derivation of Lorentz Transformation 
My doubt is in the equation (1) and (2). Aren't x,y and z also the radiuses?
EDIT
Thank you guys for trying to give a wonderful explanation but I figured out the answer myself and it was just my silly interpretation. I thought the the value of ct on x axis would be equal to ct and forgot that the value is given by a perpendicular and not an arc (of the sphere). I was thinking that where the sphere touches x axis, that is the value of ct but it isn't. The value of ct for x is given by a point on the axis that lies perpendicular to the axis.
 A: Think of the coordinates as a single object: $\mathbf{x}=\left(x,\,y,\,z\right)$. 
A point $P$ located at $\mathbf{P}=\left(a,\,b,\,c\right)$ would look like this:

Now what is the distance from the origin ($\mathbf{x}=(0,\,0,\,0)$) and $\mathbf{P}$? For this we need to use the Pythagorean theorem: the square of the diagonal is equal to the sum of the squares of the sides (e.g., $d^2=x^2+y^2+z^2$). In 2 dimensions, this is done by

$$a^2+b^2=c^2 \leftrightarrow x^2+y^2=r^2$$
But if we wanted to find the distance from some point that isn't the origin, we use
$$r^2=\left(x-x_0\right)^2+\left(y-y_0\right)^2$$
where $x,\,y$ are the positions of the point and $x_0,\,y_0$ the reference position. In the subsequent discussion, $x_0=y_0=0$.
In 3D, we can actually do the above 2D slice twice. Solve for $r$ in the $x$-$y$ plane and then solve for $d$ in the $r$-$z$ plane (using $x=a$, $y=b$, and $z=c$):
$$
r^2=a^2+b^2\to r=\sqrt{a^2+b^2}
$$
$$
d^2=r^2+c^2\to d^2=\left(\sqrt{a^2+b^2}\right)^2+c^2=a^2+b^2+c^2
$$
We could also have done this in the $y$-$z$ plane first:
$$
r^2=b^2+c^2\to r=\sqrt{b^2+c^2}
\\
d^2=r^2+a^2\to d^2=\left(\sqrt{b^2+c^2}\right)^2+a^2=a^2+b^2+c^2
$$
so clearly the ordering of it doesn't matter, we get $d=\sqrt{a^2+b^2+c^2}$ as the distance from the origin to the point $P$ when doing it both ways.
Next, what happens if we let $y=b\to y=-b$? Since we take the square of $y$, we get $y^2=\left(-b\right)^2=+b^2$ which doesn't change anything, our distance is still $d=\sqrt{a^2+b^2+c^2}$. The same thing happens if we swap the other values for their negatives. That means that there are at least 8 points $P$ that have identical distances from the origin.
Let's try using some values, just to get an idea here. Suppose we let $\mathbf{P}=\left(1,2,2\right)$. The distance $d$ is then
$$d=\sqrt{1^2+2^2+2^2}=\sqrt{1+4+4}=\sqrt{9}=3$$
If we consider a sphere with constant radius (i.e., $d$ does not change), then if we change $x=1\to x=0$ we must have that 
$$\sqrt{y^2+z^2}=3$$
 Obviously $y=z=3=d$ is not a solution because 
$$\sqrt{3^2+3^2}=\sqrt{9+9}=\sqrt{18}>4>d$$
Equally as well, $x=y=z=3=d$ is not a valid solution because
$$\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=\sqrt{27}>5>d$$
So clearly $x,\,y,\,z$ are not also the radius. Individually, the values could be the radius, but that would require the other two values being zero:
$$\sqrt{3^2+0^2+0^2}=\sqrt{9+0+0}=\sqrt{9}=3=d$$
A: The radius is the distance from the origin to any point on the sphere at time ct. If you measure this distance along the $x$ axis then the radius vector is $(ct, 0, 0)$. Likewise, measured along the $y$ axis the radius vector would be $(0, ct, 0)$ and along the $z$ axis it would be $(0, 0, ct)$. However if you measure to some point that doesn't lie on any of the three axes the radius vector would be $(x, y, z)$ where the three numbers $x$, $y$ and $z$ are related by:
$$x^2 + y^2 + z^2 = (ct)^2 $$
