SOLUTION: Give the center and radius of the circle: x^2+y^2-10x+6y+25 I first completed the square: (x^2-10x+25)+(y^2+6y+9)=-25 (x-5)^2+(y-3)^2=-25+25+9 (x-5)+(x-3)=9 (-5,-3) with

Algebra ->  Graphs -> SOLUTION: Give the center and radius of the circle: x^2+y^2-10x+6y+25 I first completed the square: (x^2-10x+25)+(y^2+6y+9)=-25 (x-5)^2+(y-3)^2=-25+25+9 (x-5)+(x-3)=9 (-5,-3) with       Log On


   

Question 214690: Give the center and radius of the circle:
x^2+y^2-10x+6y+25
I first completed the square:
(x^2-10x+25)+(y^2+6y+9)=-25
(x-5)^2+(y-3)^2=-25+25+9
(x-5)+(x-3)=9
(-5,-3) with a radius of 9 but I am not sure if I did that right?!?!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The last equation should be %28x-5%29%5E2%2B%28y%2B3%29%5E2=9, since y%5E2%2B6y%2B9 factors to %28y%2B3%29%5E2. We can take this equation and then rewrite 9 as 3%5E2 and y%2B3 to y-%28-3%29 to get %28x-5%29%5E2%2B%28y-%28-3%29%29%5E2=3%5E2


Recall that the general equation of a circle is %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 where (h,k) is the center and 'r' is the radius.


Looking at %28x-5%29%5E2%2B%28y-%28-3%29%29%5E2=3%5E2 (which is in the circle form described above), we see that h=5, k=-3 and r=3 (I'm just matching the two forms).


So the center is (5,-3) and the radius is 3 units.


So you just had some minor errors (such as the sign differences and forgetting to take the square root of 9)