Question

x^2-6x+y^2+14y+33=0 can someone please explain to me how to get the radius and center of this. i am so confused

Answer 1

you have to complete the square

Answer 2

Complete the square on : x^2-6x
Complete the square on y^2+14y

Answer 3

x^2-6x+____+y^2+14y+____=-33
x^2-6x+(1/2(6))^2+y^2+14y+(1/2(14))^2=-33+9+49
(x-3)^2+(y+7)^2=r^2 which is -33+58 which on a good day to me is 25
so the equation looks like this
\[(x-3)^2+(y+7)^2\]

Answer 4

\[(x-3)^2+(y+7)^2=25\]

Answer 5

the reason i added 9 and 49 to -33 is because thats the "completing the square" that you see on the left
(1/2(6))^2=9 and (1/2(14))^2=49
and you have to add it to the other side because its as if your adding "0" to the problem, so your allowed to do so.

Answer 6

then the center is at (k,h)= (x-k)^2+(y-h)^2=r^2
center of your problem is (3,-7) with a radius at 5

is this making sense?