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

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Answer 1 · 2017-04-14T06:05:38

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

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]}

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