Question 827709
{{{drawing(300,300,-1.1,1.3,-1.2,1.2,
circle(0,0,1),red(circle(0,0,0.02)),
green(circle(0.866,0.5,0.02)),green(circle(0.866,0,0.02)),
locate(-0.01,-0.01,O(-2,-3)),locate(0.9,0.6,P(x,y)),locate(0.85,-0.01,Q(x,-3)),
line(0,0,1,0),line(0.866,0,0.866,0.5),
line(0,0,0.866,0.5),rectangle(0.866,0,0.766,0.1)
)}}} {{{OQ=x-(-2)}}} , {{{PQ=y-(-3)}}} , and {{{OP=8}}}
According to the Pythagorean theorem for right triangle OPQ, {{{OQ^2+PQ^2=OP^2}}} so
{{{(x-(-2))^2+(y-(-3))^2=8^2}}}
{{{highlight((x+2)^2+(y+3)^2=64)}}} or we can transform it:
{{{x^2+4x+4+y^2+6y+9=64}}}
{{{x^2+y^2+4x+6y+13=64}}}
{{{x^2+y^2+4x+6y+13-64=0}}}
{{{highlight(x^2+y^2+4x+6y-21=0)}}}