Asked by gabs
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!!!
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!!!
Answers
Answered by
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^2-36) = 0
4b^2 - 8b^2 + 288=-
-4b^2 = -288
b^2 = 72
b = ± √72 or ± 6√2
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
Answered by
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 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)
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)
Answered by
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
b(b>0)- is the leg in a isosceles rectangular triangle with height=6
6^2+6^2=b^2
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