Question 389077
1.{{{x^2+y^2=36}}}
2.{{{y=x+2}}}
From eq. 2,
{{{y^2=(x+2)^2}}}
{{{y^2=x^2+4x+4}}}
Substitute into eq. 1,
{{{x^2+(x^2+4x+4)=36}}}
{{{2x^2+4x-32=0}}}
{{{x^2+2x-16=0}}}
{{{(x^2+2x+1)-16-1=0}}}
{{{(x+1)^2-17=0}}}
{{{(x+1)^2=17}}}
{{{x+1=0  +- sqrt(17)}}}
{{{highlight(x=-1 +- sqrt(17))}}}
Then from eq. 2,
{{{y=-1 +- sqrt(17)+2}}}
{{{highlight(y=1+- sqrt(17))}}}
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{{{drawing(300,300,-8,8,-8,8,grid(1),circle(-1-sqrt(17),1-sqrt(17),0.3),circle(-1+sqrt(17),1+sqrt(17),0.3),graph(300,300,-8,8,-8,8,0,sqrt(36-x^2),x+2),graph(300,300,-8,8,-8,8,0,-sqrt(36-x^2),x+2))}}}