SOLUTION: The equation x^2+y^2=169 defines a circle with its center at the origin and a radius of 13. The line y=x-7 passes through the circle. Determine the circle and line line intersect.

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: The equation x^2+y^2=169 defines a circle with its center at the origin and a radius of 13. The line y=x-7 passes through the circle. Determine the circle and line line intersect.       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 916230: The equation x^2+y^2=169 defines a circle with its center at the origin and a radius of 13. The line y=x-7 passes through the circle. Determine the circle and line line intersect.
Found 2 solutions by rothauserc, ewatrrr:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
we are given circle x^2+y^2=169 and line y = x-7
substitute for y in the circle equation using line value for y
x^2 +(x-7)^2 = 169
x^2 + x^2 -14x +49 = 169
2x^2 -14x -120 = 0
divide both sides of = by 2
x^2 -7x -60 = 0
factor polynomial
(x+5)*(x-12) = 0
x = -5 and 12
therefore the two points of intersection are
(-5, -12) and (12, 5)
to find the y value for the points just use the equation for the line

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2=169
x^2 + (x-7)^2 = 169
2x^2 -14x + 49 = 169
2x^2 -14x - 120 = 0
x^2 - 7x - 60 = 0
(x + 5)(x - 12)
x = -5, 12
and y+=+x-7
y = -12, 5
Intersection P(-5,-12) and P(12,5)