I do not understand this problem.
This is part of the question, but I think if I get this part, I'll get the rest.
Determine the value(s) of r for which the system:
(x^2)+(y^2)=(r^2)
x-2y=8
has one real solution.
I can't get anywhere! We should solve algebraically...
x^2 + y^2 = r^2 is a circle of radius r, centered at the origin
x - 2y = 8 is a line
the question is asking what r must be for the line to be tangent to the circle
the radius is perpendicular to the line at the point of tangency
... it is the shortest distance from the line to the origin (which is the center of the circle)
2y = x - 8 ... y = 1/2 x - 4
so the slope of the perpendicular is -2
... and the intercept is zero (the origin)
... y = -2 x
substituting ... -2x = 1/2 x - 4
... 4 = 2.5 x
find x, then substitute back to find y
that point is the end of the radius from the origin
... use x and y in the circle equation to find r
So my final answer would be 8/sqrt(5).
Thanks so much!
Well done, Scott.
or,
the distance from (0,0) to x-2y - 8 = 0
is
|0 - 0 - 8|/√(2^2 + 1^2)
= 8/√5
so yes, r = 8/√5
thank you, bob
we all try to bring enlightenment where we can
substituting for x, we have
(2y+8)^2 + y^2 = r^2
5y^2+32y+64-r^2 = 0
For that to have one real solution, the discriminant must be zero:
32^2-4(5)(64-r^2) = 0
r = 8/√5
Great question! Student can solve by using y in terms of x and then verify their answer by using x in terms of y.
Also like Reiny's elegant solution.