How do you find the exact solutions to the system #x^2+y^2=36# and #y=x+2#?

1 Answer
Salvatore I.
Nov 5, 2016

points:
#A(-1-root2(17),1-root2(17))# and #B(-1+root2(17),1+root2(17)#

Explanation:

If #y=x+2# this can be replaced inside the circumference so that its equation becomes
#x^2+(x+2)^2=36#
after developing the square we get the second degree equation
#2x^2+4x-32=0# that can further simplified into
#x^2+2x-16=0# that solved gives the solutions
#x_1=-1-root2(1+17)# and #x_2=-1+root2(1+16)# to which correspond the ordinates #y_1=1-root2(17)# and #y_2=1+root2(17)#

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