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!!!

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

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)

Consider the geometric method.

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

6^2+6^2=b^2

To find the values of b such that the system has one solution, we need to examine the two equations and determine the conditions under which they intersect at a single point.

First, let's rewrite the second equation in terms of x and y:
y = x + b

Now, let's substitute this expression for y in the first equation:
x^2 + (x + b)^2 = 36

Expanding and simplifying this equation leads us to:
x^2 + x^2 + 2bx + b^2 = 36
2x^2 + 2bx + b^2 = 36

Rearranging the equation:
2x^2 + 2bx + (b^2 - 36) = 0

This is a quadratic equation, and it will have one solution when the discriminant (b^2 - 4ac) is equal to zero.

The discriminant for this equation is:
Discriminant = (b^2) - (4)(2)(b^2 - 36)
= b^2 - 8b^2 + 288
= -7b^2 + 288

Setting the discriminant equal to zero:
-7b^2 + 288 = 0

Now, we solve for b:
-7b^2 = -288
b^2 = 288/7
b = ±√(288/7)

To determine the smaller and larger values for b, we can approximate this value by using a calculator or compute:

b ≈ ± 7.213464

So, the values for b are approximately:
b ≈ -7.213464 (smaller value)
b ≈ 7.213464 (larger value)

Therefore, to maintain precision, the values for b (rounded to six decimal places) are:
b ≈ -7.213464 (smaller value)
b ≈ 7.213464 (larger value)