How do you solve the system #x^2-y^2-8x+8y-24=0# and #x^2+y^2-8x-8y+24=0#?

2 Answers
Aug 11, 2016

The equations represent cencentric rectangular hyperbola of major axis #2sqrt 56# and a circle of lesser diameter #2sqrt8#, and so, there is no common point to give a solution.r

Explanation:

The first equation is

#(x-4)^2/56-(y-4)^2/56 = 1# representing a rectangular hyperbola

with center at (4, 4) and major axis #2sqrt 56#

The second has the form

(x-4)^2+(y-4)^2= 8# representing a circle with same center (4, 4) and

diameter #2sqrt 8 < 2sqrt 56#. As there is no intersection, there is

no solution..

When solved algebraically, we would be missing the graphical

aspect of the problem.

Aug 11, 2016

See below

Explanation:

Given

#{ (x^2 - y^2 - 8 x + 8 y - 24 = 0), (x^2 + y^2 - 8 x - 8 y + 24 = 0) :}#

Adding both sides we obtain

#2x^2-16x=0#. Solving for #x# we get at

#{x = 0, x = 8}#

Now substracting term to term the afore mentioned equations we obtain

#-2y^2+16y-48=0#

Solving for #y# we obtain

#y = 4 pm 2 i sqrt(2)# two complex conjugate roots. So the system dont have a real solution but has complex solutions

#( (x = 0, y = 4 - 2 i sqrt[2]), (x = 0, y = 4 + 2 i sqrt[2]), (x = 8, y = 4 - 2 i sqrt[2]), (x = 8, y = 4 + 2 i sqrt[2]))#