Question 956781
The distance from the path to (4,0) is,
{{{D[1]^2=(x-4)^2+(y-0)^2}}}
I'm assuming that by distance from {{{x=1}}} that you mean the perpendicular distance to {{{x=1}}}, that is, having no y component, only x.
The distance from the path to {{{x=1}}} is,
{{{D[2]^2=(x-1)^2+(1-1)^2}}}
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{{{2D[1]=D[2]}}}
{{{4D[1]^2=D[2]^2}}}
{{{4(x-4)^2+4y^2=(x-1)^2}}}
{{{4(x^2-8x+16)+4y^2=x^2-2x+1}}}
{{{4x^2-32x+64+4y^2=x^2-2x+1}}}
{{{3x^2-30x+63+4y^2=0}}}
{{{3(x^2-10x+21)+4y^2=0}}}
{{{3(x^2-10x+25)+21+4y^2=3(4)}}}
{{{3(x-5)^2+4y^2=12}}}
{{{highlight((x-5)^2/4+y^2/3=1)}}}
The solution is an ellipse centered at (5,0).
{{{drawing(300,300,-2,10,-6,6,grid(1),circle(4,0,.2),blue(line(1,-20,1,20)),graph(300,300,-2,10,-6,6,-sqrt((12-3(x-5)^2)/4),sqrt((12-3(x-5)^2)/4))))}}}