Fine the coordinates of points where #y=4x-8# intersects #(x-1)^2+(y-4)^2=25# ?

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Answer 1 · 2018-02-21T02:18:32

The points are #(4,8)# and #(30/17,-16/17)#

graph{(4x-8-y)((x-1)^2+(y-4)^2-25)=0 [-12.38, 12.93, -1.62, 11.04]}

Given:

#y=4x-8" [1]"#

#(x-1)^2+(y-4)^2=25" [2]"#

We can find the x coordinates of intersection by substituting equation [1] into equation [2]:

#(x-1)^2+(4x-8-4)^2=25#

#(x-1)^2+(4x-12)^2=25#

Expand the squares:

#17 x^2 - 98 x + 120 = 0#

Factor:

#(x-4)(17x-30)= 0#

#x = 4# and #x = 30/17#

Use equation [1] to find the corresponding y values:

#y = 4(4)-8#

#y = 8#

#y = 4(30/17)-8#

#y = -16/17#

The points are #(4,8)# and #(30/17,-16/17)#