Find the points of intersection of line #y=3+2x# with the circle #x^2+y^2=50#?

1 Answer
Shwetank Mauria
Apr 14, 2017

#x=-4.305# and #y=-5.61# or #x=-1.905# and #y=-0.81#

Explanation:

We have #y=3+2x# and #x^2+y^2=50#. Putting first in second, we get

#x^2+(3+2x)^2=50#

or #x^2+9+4x^2+12x=50#

or #5x^2+12x-41=0#

and using quadratic formula #x=(-12+-sqrt(144+4×5×41))/(2×5)= (-12+ sqrt964)/10#

#x=(-12+-31.05)/10#

For #-4.305# or #-1.905#

Aand #y=-4.305×2+3=-5.61# or #-1.905×2+3=-0.81#

i.e. #x=-4.305# and #y=-5.61# or #x=-1.905# and #y=-0.81#

These indicate two points at which line #y=3+2x# cuts circle #x^2+y^2=50#.

graph{(3+2x-y)(x^2+y^2-50)=0 [-20, 20, -10, 10]}

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