SOLUTION: I have an equation for a circle (x+3)^2+(y-1)^2=25 where the centre point is 3,1 and th radius is 5. I need to show that the line of equation y=-x-1 intercepts the circle. I have s

Algebra ->  Circles -> SOLUTION: I have an equation for a circle (x+3)^2+(y-1)^2=25 where the centre point is 3,1 and th radius is 5. I need to show that the line of equation y=-x-1 intercepts the circle. I have s      Log On


   

Question 28302: I have an equation for a circle (x+3)^2+(y-1)^2=25 where the centre point is 3,1 and th radius is 5. I need to show that the line of equation y=-x-1 intercepts the circle. I have substituted the value of y into the circle
(x+3)^2+(-x-1-1)^2=25
(x+3)^2+(-x-2)^2=25
But how do i get from here to 2x^2+10x-12=0 an then show that the line intercepts?
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+%28x%2B3%29%5E2%2B%28-x-2%29%5E2=25+ is fine so far. And where do you get +2x%5E2%2B10x-12=0+ from? How do you know that is the quadratic you are aiming for? :-)

Anyway, +%28x%2B3%29%5E2%2B%28-x-2%29%5E2=25+
+x%5E2+%2B+6x+%2B+9+%2B+x%5E2+%2B+4x+%2B+4+=+25+
+2x%5E2+%2B+6x+%2B+9+%2B+4x+%2B+4+=+25+
+2x%5E2+%2B+10x+%2B+9+%2B+4+=+25+
+2x%5E2+%2B+10x+%2B+13+=+25+
+2x%5E2+%2B+10x+-+12+=+0+
+x%5E2+%2B+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.