pre calc

posted by .

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 -

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^2-36) = 0
4b^2 - 8b^2 + 288=-
-4b^2 = -288
b^2 = 72
b = ± √72 or ± 6√2

• pre calc -

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 y-intercept 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^2-36=0

Since the line is tangent then the equation has unique solution =>

the discriminant (2b)^2-4*2(b^2-36)=0
4b^2-8b^2+288=0
b^2=72
b=6sqrt(2) or b=-6sqrt(2)

• pre calc -

Consider the geometric method.

b(b>0)- is the leg in a isosceles rectangular triangle with height=6

6^2+6^2=b^2