Question 714400
The intercepts will be when x=0 and you solve for y, and when y=0 and you solve for x.  Both situations will give you quadratic equation in one variable. 


{{{0*0+y^2+10y-24=0}}} becomes {{{y^2+10y-24=0}}} for whe x=0.  Easily factorable to {{{(y-2)(y+12)=0}}}, so intercepts will be at y=2 and y=-12.  So see how that works?  You can do the same for finding the intercepts when y=0.


Finding radius and center need a different equation form.  Put the equation into standard form by Completing the square.  Do this for BOTH variables.


Here's getting the process started.
x^2+y^2+10y-24=0
x is already in good shape.  No need to complete the square for x.  y needs some work.
The square term to add and subtract is {{{(10/2)^2=25}}}.
{{{(y^2+10y+25)-25=(y+5)^2-25}}}.  We now must include that in the original equation:


Putting that result into the original equation, we obtain:
{{{x^2+ (y+5)^2-25 -24 =0}}}
{{{highlight(x^2+(y+5)^2=49)}}}
OR
{{{highlight(x^2+(y+5)^2=7^2)}}}


Center is at (0,-5) and radius is 7.