How do you solve #(x-3)^2+(y-5)^2=25# and #x^2+(y-1)^2=100#?

1 Answer
Aug 8, 2015

Answer:

Expand the squares for the 2 equations for circles, subtract to get a linear equation, substitute back into one of the circle equations:
#color(white)("XXXX")##(x,y) = (6,9)#

Explanation:

[1]#color(white)("XXXX")##(x-3)^2+(y-5)^2=25#
[2]#color(white)("XXXX")##x^2+(y-1)^2=100#

Expanding the square for [1]
[3]#color(white)("XXXX")##x^2-6x+9 + y^2-10y+25 = 25#
Simplifying
[4]#color(white)("XXXX")##x^2-6x+y^2-10y = -9#

Expanding the square for [2]
[5]#color(white)("XXXX")##x^2 +y^2-2y+1 = 100#
Simplifying
[6]#color(white)("XXXX")##x^2+y^2-2y= 99#

Subtract [4] from [6]
[7]#color(white)("XXXX")##6x +8y = 108#
Simplify
[8]#color(white)("XXXX")##y = (54-3x)/4#

Substituting #((54-3x)/4)# for #y# in [2]
[9]#color(white)("XXXX")##x^2+((54-3x)/4 -1)^2 = 100#
or
[10]#color(white)("XXXX")##x^2+(50/4-(3x)/4)^2 = 100#

Expanding the square
[11]#color(white)("XXXX")##x^2 + (2500/16-(300x)/16+(9x^2)/16) = 100#

Multiply by 16
[12]#color(white)("XXXX")##16x^2 +2500-300x+9x^2 = 1600#

Simplifying into standard quadratic form
[13]#color(white)("XXXX")##25x^2-300x+900 = 0#
Dividing by #25#
[14]#color(white)("XXXX")##x^2-12x+36 = 0#

Factoring
[15]#color(white)("XXXX")##(x-6)^2 = 0##color(white)("XXXX")##rarr x=6#

When #x=6# (substituting back into [8]

[16]#color(white)("XXXX")##y = (54-3(6))/4 = 9#