Question 138687
{{{x^2+y^2=26}}} Start with the given equation



{{{y=0+-sqrt(26-x^2)}}} Solve for y



{{{y=sqrt(26-x^2)}}} or {{{y=-sqrt(26-x^2)}}} Break up the right side




Now let's graph {{{sqrt(26-x^2)}}}


{{{ graph( 500, 500, -10, 10, -10, 10, sqrt(26-x^2) ) }}} Graph of {{{y=sqrt(26-x^2)}}}



Now let's graph {{{-sqrt(26-x^2)}}}


{{{ graph( 500, 500, -10, 10, -10, 10, -sqrt(26-x^2) ) }}} Graph of {{{y=-sqrt(26-x^2)}}}



Now graph {{{y=-x+8}}}


{{{ graph( 500, 500, -10, 10, -10, 10, -x+8  ) }}} Graph of {{{y=-x+8}}}



Now graph the three equations together (note: the first two equations make up {{{x^2+y^2=26}}} )


{{{ graph( 500, 500, -10, 10, -10, 10, sqrt(26-x^2),-sqrt(26-x^2),-x+8  ) }}} Graph of {{{y=sqrt(26-x^2)}}} (red). Graph of {{{y=sqrt(26-x^2)}}} (green). Graph of {{{y=-x+8}}}(blue)



From the graph, we can see that there are no intersections. So there are no solutions.