A circle of radius r1 is inscribed in the first quadrant such that it is tangent to the line x = 10 and y = 10, and also externally tangent to the circle x² + y² = 100 (radius 10). The radius of the smaller circle can be expressed as r1 = a - b*sqrt(2). Find (a - b)/2.

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1. The small circle touches x=10 and y=10, so its centre is at (10 - r1, 10 - r1). For external tangency with x²+y²=100 (centre at origin, radius 10), the distance between centres = 10 + r1. So sqrt((10-r1)² + (10-r1)²) = 10 + r1. This gives sqrt(2)*(10-r1) = 10+r1. So 10*sqrt(2) - r1*sqrt(2) = 10 + r1. Thus r1*(1+sqrt(2)) = 10*sqrt(2) - 10 = 10*(sqrt(2)-1). So r1 = 10*(sqrt(2)-1)/(sqrt(2)+1). Rationalise: multiply by (sqrt(2)-1)/(sqrt(2)-1): r1 = 10*(sqrt(2)-1)²/(2-1) = 10*(2 - 2*sqrt(2) + 1) = 10*(3 - 2*sqrt(2)) = 30 - 20*sqrt(2). So a=30, b=20, and (a-b)/2 = (30-20)/2 = 10/2 = 5. Or if a=30, b=20, then a-b=10, (a-b)/2=5. Hmm, that's 5 not 1. Checking options: perhaps the answer should be

A circle of radius r1 is inscribed in the first quadrant such that it is tangent to the line x = 10 and y = 10, and also externally tangent to the circle x² + y² = 100 (radius 10). The radius of the smaller circle can be expressed as r1 = a - b*sqrt(2). Find (a - b)/2.

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