Posted by gabs on .
Find the values of b such that the system has one solution.
x^2 + y^2 = 36
y= x+b
and then they give me an answer box
b=_________ (smaller value)
b=_________ (larger value)
i have absolutely no idea how to even begin this and what they are asking. could some please help and explain what im supposed to do with this. thanks so much!!!

pre calc 
Reiny,
sub in the straight line into the circle
x^2 + (x+b)^2 = 36
x^2 + x^2 + 2bx + b^2  36 = 0
for one solution the discriminant, that is
b^2  4ac = 0
(2b)^2  4(2)(b^236) = 0
4b^2  8b^2 + 288=
4b^2 = 288
b^2 = 72
b = ± √72 or ± 6√2 
pre calc 
Mgraph,
Draw two tangents to the circle x^2+y^2=6^2 parallel to the line y=x in the second and fourth quadrants.
b is yintercept of these tangents.
Сonsider the algebraic method.
Find the coordinates of common points of the line and circle:
x^2+(x+b)^2=36
2x^2+2bx+b^236=0
Since the line is tangent then the equation has unique solution =>
the discriminant (2b)^24*2(b^236)=0
4b^28b^2+288=0
b^2=72
b=6sqrt(2) or b=6sqrt(2) 
pre calc 
Mgraph,
Consider the geometric method.
b(b>0) is the leg in a isosceles rectangular triangle with height=6
6^2+6^2=b^2