document.write( "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. )\r
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document.write( "Show all work and give an exact answer - no rounding at any time, and no decimals.\r
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document.write( "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. \n" );
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Algebra.Com's Answer #450193 by Alan3354(69443) 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. \n" ); document.write( "================ \n" ); document.write( "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. \n" ); document.write( "---------- \n" ); document.write( "Use the x & y axes \n" ); document.write( "The larger circle is (x-1)^2 + (y-1)^2 = 1 \n" ); document.write( "The tangent point (call it T) on both circles is (1 - sqrt(2)/2,1 - sqrt(2)/2) [apx (0.2928,0.2928)] \n" ); document.write( "--------------- \n" ); document.write( "The center of the small circle is equidistant from T, the x-axis and the y-axis. \n" ); document.write( "For T at (x,y): \n" ); document.write( " \n" ); document.write( "Since x = y \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "-------------------- \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "---------- \n" ); document.write( "Now it's a quadratic. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "=================== \n" ); document.write( "A mistake somewhere, see if you can find it.\r \n" ); document.write( "\n" ); document.write( "I'll finish it later, but this is how it's done. \n" ); document.write( " |