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.
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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):
{{{x = y = sqrt((x - 1 + sqrt(1/2))^2 + (y - 1 + sqrt(1/2))^2)}}}
Since x = y
{{{x = sqrt((x - 1 + sqrt(1/2))^2 + (x - 1 + sqrt(1/2))^2)}}}
{{{x = sqrt(2(x - 1 + sqrt(1/2)))^2}}}
{{{x = sqrt(2*(x^2 + 1 + 2x*sqrt(1/2) - 2sqrt(1/2) + 1/2))}}} *** I left out a term of -2x
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{{{x^2 = 2*(x^2 + 2x*sqrt(1/2) - 2sqrt(1/2) + 3/2)}}}
{{{x^2 = 2x^2 + 4x*sqrt(1/2) - 4sqrt(1/2) + 3}}}
{{{x^2 + (4x - 4)*sqrt(1/2) + 3 = 0}}}
{{{x^2 + x*sqrt(8) - sqrt(8) + 3 = 0}}}
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Now it's a quadratic.
{{{x = -sqrt(2) + sqrt((8 - 4*(3 - sqrt(8))))/2 }}}
{{{x = -sqrt(2) + sqrt(8 - 12 + 4sqrt(8))/2 }}}
{{{x = -sqrt(2) + sqrt(4sqrt(8) - 4)/2 }}}
{{{x = -sqrt(2) + sqrt(sqrt(8) - 1)}}}
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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.