How do you find the intersection between #y=x-8# and #x^2+y^2= 34#?

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Answer 1 · 2016-11-25T21:32:58

Points of intersection are:

#(x_1,y_1) = (3 ,-5)#

#(x_2,y_2)=(5,-3)#

#x^2+y^2=34# is the equation of a circle or radius #sqrt(34)# and has its centre at the origin.

Expressing this with #y# being the dependant variable we have:

#y=+-sqrt(34-x^2)# ...................Equation(1)

We also have the straight line equation

#y=x-8#......Equation(2)

substitute for #y# in equation(1) using equation(2) we have:

#x-8=+-sqrt(34-x^2)#

Squaring both sides

#(x-8)^2=34-x^2#

#x^2-16x+64=34-x^2#

#2x^2-16x+30=0#

Divide both sides by 2

#x^2-8x+15=0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#x=(-(-8)+-sqrt( (-8)^2-4(1)(15)))/(2(1))#

#x=4+-1#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using equation(2)

At #x=3 -> y=x-8 = -5#

At #x=5->y=x-8=-3#

Points of intersection are:

#(x_1,y_1) = (3 ,-5)#

#(x_2,y_2)=(5,-3)#

Tony B