Posted by **gabs** on Tuesday, June 7, 2011 at 10:56pm.

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**, Tuesday, June 7, 2011 at 11:20pm
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 -
**Mgraph**, Tuesday, June 7, 2011 at 11:37pm
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 -
**Mgraph**, Tuesday, June 7, 2011 at 11:52pm
Consider the geometric method.

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

6^2+6^2=b^2

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