Question 759761
{{{drawing(300,300,-1.9,1.9,-1.9,1.9,
grid(0),circle(0,0,1),
blue(triangle(-1.8,1.35,-1.8,0,0,0)),
blue(line(-0.8,0.6,-0.8,0)),
blue(rectangle(-1.8,0,-1.7,0.1)),
blue(rectangle(-0.8,0,-0.7,0.1)),
locate(-0.8,0.8,B),locate(-1.9,1.6,A(-x,y)),
locate(-0.9,1.45,x),arrow(-0.94,1.35,-1.78,1.35),
arrow(-0.78,1.35,-0.02,1.35),
locate(-0.8,-0.05,Q),locate(-1.8,-0.05,P),
locate(0.05,0.2,O),locate(-1.75,0.8,y)
)}}} I constructed OAP and OBQ as right triangles.
{{{AP=y}}} and {{{OP=x}}}
The Pythagorean theorem says that {{{OA=sqrt(x^2+y^2)}}}
Since OAP and OBQ have the same angle at O, and a {{{90^o}}} angle, they are similar right triangles.
Since they are similar, corresponding sides are proportional.
So {{{OQ/OB=OP/OA}}} and {{{BQ/OB=AP/OA}}}
We know the lengths of the sides of OAP (AP, OP, and OA).
We know {{{OB=1}}} because it is the radius of the unit circle.
We can find the length of sides OQ and BQ:
{{{OQ/OB=OP/OA}}} means {{{OQ/1=x/sqrt(x^2+y^2)}}} --> {{{OQ=x/sqrt(x^2+y^2)}}}
{{{BQ/OB=AP/OA}}} means {{{BQ/1=y/sqrt(x^2+y^2)}}} --> {{{BQ=y/sqrt(x^2+y^2)}}}
The x-coordinate of B is {{{-OQ=highlight(-x/sqrt(x^2+y^2))}}}.
The y-coordinate of B is {{{BQ=highlight(y/sqrt(x^2+y^2))}}}.