Question 846641
Intersection of a circle and a straight line.
Substitute,
{{{y=2-x}}}
{{{x^2+(2-x)^2=9}}}
{{{x^2+(4-4x+x^2)=9}}}
{{{2x^2-4x+4=9}}}
{{{2x^2-4x-5=0}}}
Use quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-(-4) +- sqrt( (-4)^2-4*2*(-5) ))/(2*2) }}}
 {{{x = (4+- sqrt( 16+40))/(4)}}}
{{{x = (4+- sqrt( 56))/(4)}}}
{{{x = (4+- 2*sqrt( 14))/(4)}}}
{{{x = 1+- sqrt( 14)/2}}}
 
Then,
{{{y=2-x}}}
{{{y=2-(1+-sqrt(14)/2))}}}
{{{y=1 +- sqrt(14)/2}}} Remember the +/- sign should be -/+ but the symbol doesn't exist here.

({{{1-sqrt(14)/2}}},{{{1+sqrt(14)/2}}})
and
({{{1+sqrt(14)/2}}},{{{1-sqrt(14)/2}}})
.
.
.
{{{drawing(300,300,-5,5,-5,5,grid(1),circle(0,0,3),
circle(1-sqrt(14)/2,1+sqrt(14)/2,0.15),
circle(1+sqrt(14)/2,1-sqrt(14)/2,0.15),
graph(300,300,-5,5,-5,5,2-x))}}}