How do you solve #x^2+y^2+6y+5=0# and #x^2+y^2-2x-8=0#?

1 Answer
Sep 20, 2015

Answer:

Resolve to a quadratic in #x# and solve to find:

#(x, y) = ((5+-sqrt(1215))/20, -(135+-sqrt(1215))/60)#

(with matching signs for #+-#)

Explanation:

Subtract the second equation from the first to get:

#6y + 2x + 13 = 0#

Subtract #2x + 13# from both sides to get:

#6y = -(2x+13)#

Divide both side by #6# to get:

#y = -(2x+13)/6#

Substitute this expression for #y# in the first equation to get:

#0 = x^2 + (2x+13)^2/36 -(2x+13)+5#

#= x^2 + (2x+13)^2/36-2x-8#

Multiply through by #36# to get:

#0 = 36x^2 + (2x+13)^2 -72x - 288#

#=36x^2 + 4x^2+52x+169-72x-288#

#=40x^2-20x-119#

Solve this using the quadratic formula to get:

#x = (20 +-sqrt(20^2-(4xx40xx-119)))/(2xx40)#

#= (20 +- sqrt(400+19040))/80#

#=(20 +- sqrt(19440))/80#

#=(5+-sqrt(1215))/20#

So:

#y = -(2((5+-sqrt(1215))/20)+13)/6#

#=-(20((5+-sqrt(1215))/20)+130)/60#

#=-(5+-sqrt(1215)+130)/60#

#=-(135+-sqrt(1215))/60#

graph{(x^2+y^2+6y+5)(x^2+y^2-2x-8) = 0 [-9.755, 10.245, -6.04, 3.96]}