SOLUTION: A large circle of radius 1 is drawn tangent to perpendicular rays. Find the radius of the biggest possible circle that is placed in the space. (The smaller circle must also be tang

Question 737118: A large circle of radius 1 is drawn tangent to perpendicular rays. Find the radius of the biggest possible circle that is placed in the space. (The smaller circle must also be tangent to the perpendicular rays. )
Show all work and give an exact answer - no rounding at any time, and no decimals.
Remember: the radius of the small circle does not include the space between the small circle and the right angle formed by the perpendicular rays.
Found 2 solutions by josgarithmetic, Alan3354:
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
Smallest or otherwise, your description is only one particular circle. It can easily be drawn according to how you describe. The radius is 1 unit. Imagine this circle in the cartesian plane, centered at (1, 1) which will be tangent to the two cartesian axes at (1, 0) and at (0, 1).

You ask for steps to finding the "largest " circle tangent to two perpendicular rays, such that the circle is radius 1. This is only one circle. Using standard form, filling in the information as if done on cartesian system, one possible circle is then (x-1)^2+(y-1)^2=1, and the perpendicular rays are the x and y axes in their positive directions meeting with shared endpoints at the origin. No other steps really.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A large circle of radius 1 is drawn tangent to perpendicular rays. Find the radius of the biggest possible circle that is placed in the space.
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You didn't make it clear, but I think you mean a smaller circle tangent to the 2 rays and tangent to the larger circle.
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Use the x & y axes
The larger circle is (x-1)^2 + (y-1)^2 = 1
The tangent point (call it T) on both circles is (1 - sqrt(2)/2,1 - sqrt(2)/2) [apx (0.2928,0.2928)]
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The center of the small circle is equidistant from T, the x-axis and the y-axis.
For T at (x,y):

Since x = y

*** I left out a term of -2x
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Now it's a quadratic.

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A mistake somewhere, see if you can find it.
I'll finish it later, but this is how it's done.
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