SOLUTION: Find the center, radius, and intercepts of the circle with the given equation x^2+y^2+10y-24=0

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Question 714400: Find the center, radius, and intercepts of the circle with the given equation x^2+y^2+10y-24=0
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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%2A0%2By%5E2%2B10y-24=0 becomes y%5E2%2B10y-24=0 for whe x=0. Easily factorable to %28y-2%29%28y%2B12%29=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 %2810%2F2%29%5E2=25.
%28y%5E2%2B10y%2B25%29-25=%28y%2B5%29%5E2-25. We now must include that in the original equation:

Putting that result into the original equation, we obtain:
x%5E2%2B+%28y%2B5%29%5E2-25+-24+=0
highlight%28x%5E2%2B%28y%2B5%29%5E2=49%29
OR
highlight%28x%5E2%2B%28y%2B5%29%5E2=7%5E2%29

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