Find the center and radius of the circle given by
x^2-8x+y^2+6y-39=0

Question

Find the center and radius of the circle given by
x^2-8x+y^2+6y-39=0

anonymous · Answer

You need to complete the square of the two binomials separately. 

x^2-8x and y^2+6y

anonymous · Answer

Do you know how to do that?

anonymous · Answer

no

anonymous · Answer

Ok, 

take the number in front of the variable with no exponent, divide it by two, and then square it. Then add and subtract that number to the end of the binomial.

x^2-8x+16-16+y^2+6y+9-9-39

Factor the two trinomials and move the left over numbers to the right

(x-4)^2+(y+3)^2=64

anonymous · Answer

Now you have it in the right form. 

(x-h)^2+(y-k)^2=r^2

h is the x coordinate of the center, k is the y coordinate of the center, and r is radius. Can you see what they are now?

anonymous · Answer

So the center would be (4,-3) and the radius would be square root of 64 which is 8.

anonymous · Answer

Thanks. I appreciate it much.