how does this translate to a circle with radius 5: $\sqrt{24-2x-x^2}$ I tried squaring both sides to get this $y^2 = 24-2x-x^2$, then putting the $x$'s with the $y$'s to get $y^2 + x^2 + 2x = 24$. Then I tried dividing everything by 24, but I don't see it. Tried factoring too.
 A: $$y=\sqrt{24-2x-x^2}\\y^2=24-2x-x^2\\y^2=25-(x+1)^2\\(x+1)^2+y^2=5^2$$
From the second to the third line, I completed the square.  Notice now that we have the equation of a circle centered at $(-1,0)$ of radius $5$.  However, your original equation only represents the top half of this circle, because $y$ is always positive.
Here is a picture of the graph:

A: equation of circle in general form is 
$$  x^2 + y^2 + (2\times g \times x) + (2\times f \times y) + c = 0 $$ 
$$\boxed{Radius =  \sqrt{g^2 + f^2 -c }}$$  
compare this with your equation 
$$   x^2 + y^2 + (2\times 1 \times x) + (2\times 0 \times y) + (-24) = 0 $$ 
here g = 1 , f =0 , c = -24 
so radius = $ \sqrt { 1^2 +0^2 -(-24)} $
= $\sqrt {25 } $
= 5
From above way  the tedious job of factorizing is eliminated .
The equation given in your case is equation in general form
Why I suggested this solution is that sometimes it is tedious to change a equation from general form $  x^2 + y^2 + (2\times g \times x) + (2\times f \times y) + c = 0 $ , to standard form $ (x-a)^2 + (y-b)^2 = (radius)^2 $, in those cases the radius formula given above will be helpful .
A: the equation was; x^2 + y^2 + 2x -24=0   is in the general form,
and the general equation of the circle is ax^2 + ay^2 + dx + ey + f=0;
using the equation alone we can solve for the radius,center,area and circum.||
Using the formulas.:||
center:    h= -d/2a  ,  k=-e/2a||
radius;    r^2= (d^2+ e^2 -4af)/4a^2||
area ;         A= pi(r^2)||
applying:||
given: a=1    d=2   e=0   f=-24||
center;  h=(-2)/2(1)=-1      k=-(0)/2(1)=0||
radius;  r^2=(2^2 + 0^2 - 4(1)(-24))/4(1^2)||
                  r^2=25  = 5  ||
