Question 913984
A circle of radius and centered at the origin and passes through the point {{{P(1/2,sqrt(3)/2)}}} .
 
The equation for a line that starts with that starts with "y =" is the equation in slope-intercept form,
{{{y=m*x+b}}}, where {{{m}}} is the slope and {{{b}}} is the y-intercept.
 
(a) The line through the origin, {{{O(0,0)}}} , and the point {{{P(1/2,sqrt(3)/2)}}}
has a y-intercept of {{{b=0}}} , and 
a slope of {{{m=(sqrt(3)/2-0)/(1/2-0)=(sqrt(3)/2)/(1/2)=(sqrt(3)/2)*(2/1)=sqrt(3)}}}

The equation for the line hrough the origin and the point {{{P(1/2,sqrt(3)/2)}}} that starts with "y =" is
{{{highlight(y= sqrt(3)x)}}} .
 
(b) The tangent line to the circle at the point {{{P(1/2,sqrt(3)/2)}}} goes through the point of tangency, {{{P(1/2,sqrt(3)/2)}}} .
The radius of the circle at the point of tangency is the line segment that connects the points {{{O(0,0)}}} and {{{P(1/2,sqrt(3)/2)}}} .
That line segment is part of the line with equation {{{y= sqrt(3)x}}} and slope {{{m=sqrt(3)}}} found in part a).
According to the hint, the tangent you are looking for is perpendicular to that line and passes through the point of tangency, {{{P(1/2,sqrt(3)/2)}}} .
The line perpendicular to a line with slope {{{m}}}, has a slope of {{{-1/m}}} .
So, for the tangent you are looking for,
the slope is {{{-1/sqrt(3)}}} .
That is more elegantly expressed as
{{{(-1/sqrt(3))*(sqrt(3)/sqrt(3))=-sqrt(3)/3}}}
The equation of that tangent in point-slope form, based on point {{{P(1/2,sqrt(3)/2)}}} is
{{{highlight(y-sqrt(3)/2=(-sqrt(3)/3)(x-1/2))}}} .
If you wanted to express that equation in slope intercept form, you would "distribute" and "solve for y", like this:
{{{y-sqrt(3)/2=(-sqrt(3)/3)(x-1/2)}}}--->{{{y-sqrt(3)/2=(-sqrt(3)/3)x-(1/2)(-sqrt(3)/3)}}}--->{{{y-sqrt(3)/2=(-sqrt(3)/3)x+sqrt(3)/6}}}--->{{{y=(-sqrt(3)/3)x+sqrt(3)/6+sqrt(3)/2}}}--->{{{y=(-sqrt(3)/3)x+sqrt(3)/6+3sqrt(3)/6}}}--->{{{y=(-sqrt(3)/3)x+4sqrt(3)/6}}}--->{{{highlight(y=(-sqrt(3)/3)x+2sqrt(3)/3)}}}