Question

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

Answer 1

`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]}