SOLUTION: Could you please help me solve this? find the center (h,k) and radius r of the circle. I have no idea how to work out. x^2+y^2-12x+6y+35=0

Algebra ->  Systems-of-equations -> SOLUTION: Could you please help me solve this? find the center (h,k) and radius r of the circle. I have no idea how to work out. x^2+y^2-12x+6y+35=0      Log On


   



Question 126884: Could you please help me solve this? find the center (h,k) and radius r of the circle. I have no idea how to work out.
x^2+y^2-12x+6y+35=0

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Start with the general formula for a circle
%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2
Somehow you have to get your equation into that form
x%5E2+%2B+y%5E2+-+12x+%2B+6y+%2B+35+=+0
Group the x terms together and the y terms together
and subtract 35 from both sides
x%5E2+-+12x+%2B+y%5E2+%2B+6y+=+-35
Now do the operation called completing the square
on the x terms and the y terms
You do that by taking 1/2 of the coefficient of
x (or y), squaring it, then add it to both sides

Notice that this is the exact same equation
I started with because I added the same thing to BOTH sides
x%5E2+-+12x+%2B+36+%2B+y%5E2+%2B+6y+%2B+9+=+-35+%2B+36+%2B+9
Now you can get this into the form
%28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2
%28x+-+6%29%5E2+%2B+%28y+%2B+3%29%5E2+=+10
If you just match up the terms, you see that
h+=+6
k+=+-3
r%5E2+=+10
r+=+sqrt%2810%29
So, the center is at (6,-3) and the radius is sqrt%2810%29