Question 217702
{{{x^2+y^2=1}}}  This is an equation of a circle centered at point (0,0) with radius 1.


{{{x+y=-1}}}  This is an equation of a line with slope m=-1 and y-intercept at b=-1 or at point (0,-1).


Step 1.  Let's solve for x in {{{x+y=-1}}}


Subtract -y to both sides of the equation


{{{x+y-y=-1-y=-1(1+y)=-1(y+1)}}}}


{{{x=-1(y+1)}}}


Step 2.  Substitute {{{x=-1(y+1)}}} into {{{x^2+y^2=1}}}


{{{x^2+y^2=1= (-1)^2(y+1)^2+y^2}}}  but {{{(-1)^2=1}}} and {{{(y+1)^2=y^2+2y+1}}}


{{{x^2+y^2=y^2+2y+1+y^2=1}}}


{{{2y^2+2y+1=1}}}


Step 3.  Subtract 1 to both sides of the equation


{{{2y^2+2y+1-1=1-1}}}


{{{2y^2+2y=0}}}


{{{2y(y+1)=0}}}


Then {{{y=0}}} and {{{y=-1}}}


Step 4.  Find x-coordinates when {{{y=0}}} and {{{y=-1}}}


At {{{y=0}}}, then {{{x=-1-y=-1-0=-1}}}.  One intersection point is (-1,0).


At {{{y=-1}}} then {{{x=-1-(-1)=-1+1=0}}}.  One intersection point is (0,-1).


Step 5.  Then intersection points are (-1,0) and (0,-1)


Here are the graphs of the circle and line and note they have two common points.





{{{graph(300,300, -1, 1, -1, 1, -x-1, sqrt(1-x^2), -sqrt(1-x^2))}}}


I hope the above steps were helpful. 


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


And good luck in your studies!


Respectfully,
Dr J