document.write( "Question 969255: Write the equation of the circle satisfying the given condition.
\n" );
document.write( "Radius of 2, tangent to y=2x and passing through (3,-4) \n" );
document.write( "
Algebra.Com's Answer #592191 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! Write the equation of the circle satisfying the given condition. \n" ); document.write( "Radius of \n" ); document.write( "=============== \n" ); document.write( "The center has to be on the line parallel to y = 2x and \n" ); document.write( "Find the line's equation: \n" ); document.write( "y = 2x has a slope of 2 and passes thru (0,0) \n" ); document.write( "A line perpendicular to it (RS) thru (0,0) is y = -x/2 \n" ); document.write( "A point on the line RS sqrt(20) from (0,0) is the intersection of a circle centered at (0,0) with r = sqrt(20) \n" ); document.write( "--> x^2 + y^2 = 20 \n" ); document.write( "y = -x/2 \n" ); document.write( "x^2 + x^2/4 = 20 \n" ); document.write( "x = 4 (Ignore the x = -4 in Q2, (3,-4) is in Q4) \n" ); document.write( "--> (4,-2) \n" ); document.write( "The center of the circle is on a line thru (4,-2) with m = 2 (parallel to y = 2x) \n" ); document.write( "y+2 = 2(x-4) = 2x - 8 \n" ); document.write( "y = 2x - 10 \n" ); document.write( "---- \n" ); document.write( "Find the point(s) sqrt(20) from (3,-4) on the line y = 2x - 10 \n" ); document.write( "It's the intersections of a circle centered at (3,-4) r = sqrt(20) \n" ); document.write( "--- \n" ); document.write( " \n" ); document.write( "Sub for y \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "x = 5 --> center @ (5,0) --> \n" ); document.write( "---- \n" ); document.write( "x = 1 --> center @ (1,-8) --> \n" ); document.write( "============================= \n" ); document.write( "There must be an easier way to do this, but I don't know of one. \n" ); document.write( " |