Question 28302
{{{ (x+3)^2+(-x-2)^2=25 }}} is fine so far. And where do you get {{{ 2x^2+10x-12=0 }}} from? How do you know that is the quadratic you are aiming for? :-)


Anyway, {{{ (x+3)^2+(-x-2)^2=25 }}}
{{{ x^2 + 6x + 9 + x^2 + 4x + 4 = 25 }}}
{{{ 2x^2 + 6x + 9 + 4x + 4 = 25 }}}
{{{ 2x^2 + 10x + 9 + 4 = 25 }}}
{{{ 2x^2 + 10x + 13 = 25 }}}
{{{ 2x^2 + 10x - 12 = 0 }}}
{{{ x^2 + 5x - 6 = 0 }}}
(x+6)(x-1) = 0


so x+6=0 OR x-1=0
--> x=-6 OR x=1


This says there are 2 solutions of the "circle" and the "line" ie where they are EQUAL ie where they intersect.


I have told you their x-coordinates. If you need to quote the y values....find them too, from y=-x-1... easiest equation to use.


jon.