SOLUTION: how do I graph these two equations ? (x^2-12x+84=y^2+16y) , (x^2+y^2=4x+9) ? (ie, simplify, get center/radius)

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Question 390094: how do I graph these two equations ? (x^2-12x+84=y^2+16y) , (x^2+y^2=4x+9) ? (ie, simplify, get center/radius)
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
If you want to simply graph these two equations, solve for y then plot it manually or on a graphing calculator.
given: x^2-12x+84=y^2+16y
complete the square
x^2-12x+36-( y^2+16x+64)=-84+36-64=-112
(x-6)^2-(y+8)^2=-112
multiply by (-1)
(y+8)^2-(x-6)^2=112
(y+8)^2/112 + (x-6)^2/112 = 1
This is a hyperbola with center at (6,-8), and transverse axis in the vertical direction.
solving for y
(y+8)^2 = 112+(x-6)^2 = 112+x^2-12x+36
y+8 =sqrt( 112+x^2-12x+36) =( 112+x^2-12x+36)^.5
y=-8+( 112+x^2-12x+36)^.5, and -8-( 112+x^2-12x+36)^.5
The first expression will graph the top half of the hyperbola and the second, the bottom half
Now, for the other given equation, x^2+y^2=4x+9
complete the square
x^2-4x+4+y^2 = 9+4=13
(x-2)^2+y^2 = 13
This is a circle with center at (2,0) with radius sqrt(13)
solving for y
y^2 = 13-(x-2)^2=13-x^2+4x-4
y = ħsqrt(13-x^2+4x-4) or ħ(13-x^2+4x-4)^.5
The first (+)expression will graph the top half of the circle and the second(-), the bottom half
The following graph will give you some idea what the curves look like.