SOLUTION: Find the center and the radius of the circle 3x^2+3y^2+15x-11y-10=0 algebraically,

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Question 1024402: Find the center and the radius of the circle 3x^2+3y^2+15x-11y-10=0 algebraically,
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2%2B3y%5E2%2B15x-11y-10=0

We need to get it in the form %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 so we'll
know it has center (h,k) and radius r.

First we add 10 to both sides to get the constant off the left side:

3x%5E2%2B3y%5E2%2B15x-11y=10

Next we divide through by the common coefficient of x² and y² which
is 3

expr%283%2F3%29x%5E2%2Bexpr%283%2F3%29y%5E2%2Bexpr%2815%2F3%29x-expr%2811%2F3%29y=10%2F3

x%5E2%2By%5E2%2B5x-expr%2811%2F3%29y=10%2F3

Swap the 2nd and 3rd terms to get the terms in like letters together: 

x%5E2%2B5x%2By%5E2-expr%2811%2F3%29y=10%2F3

Put the addition of one blank after the 5x and another after the
expr%28-11%2F3%29y since we are going to add something to both sides

x%5E2%2B5x%2B%22__%22%2By%5E2-expr%2811%2F3%29y%2B%22__%22=10%2F3

Next we calculate what goes in the first blank:

1. Multiply the coefficient of x, which is 5, by 1%2F2}, getting 5%2F2
2. Square that value:  %285%2F2%29%5E2+=+25%2F4, so red%2825%2F4%29 is what goes
in the first blank, and we also add it to the right side:

x%5E2%2B5x%2Bred%2825%2F4%29%2By%5E2-expr%2811%2F3%29y%2B%22__%22=10%2F3%2Bred%2825%2F4%29  

Next we calculate what goes in the second blank:

1. Multiply the coefficient of y, which is -11%2F3, by 1%2F2}, getting -11%2F6
2. Square that value:  %28-11%2F6%29%5E2+=+121%2F36, so red%28121%2F36%29 is 
what goes in the second blank, and we also add it to the right side:



The first three terms on the left x%5E2%2B5x%2B25%2F4 factors as %28x%2B5%2F2%29%28x%2B5%2F2%29 or %28x%2B5%2F2%29%5E2

The last three terms on the left y%5E2-expr%2811%2F3%29y%2B121%2F36 factors as %28y-11%2F6%29%28y-11%2F6%29 or %28y-11%2F6%29%5E2

The right side 10%2F3%2B25%2F4%2B121%2F36 becomes 120%2F36%2B225%2F36%2B121%2F36=466%2F36=233%2F18

So we now have:

%28x%2B5%2F2%29%5E2%2B%28y-11%2F6%29%5E2=233%2F18

And we compare that to 
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 

and we see that since -h=%22%22%2B5%2F2%22, h=-5%2F2, and
since -k=-11%2F6, k=11%2F6. Also r%5E2=233%2F18 so r=sqrt%28233%2F18%29

If you like you can rationalize r=sqrt%28expr%28233%2F18%29%2Aexpr%282%2F2%29%29=sqrt%28466%2F36%29=sqrt%28466%29%2F6 

Edwin