document.write( "Question 1064393: Find the equation of the circle satisfying the given conditions (general equation)\r
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document.write( "1. Tangent to the line 4x-3y=6 at (3,2) and passing through (2,-1). \n" );
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Algebra.Com's Answer #679448 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! Find the equation of the circle satisfying the given conditions (general equation)\r \n" ); document.write( "\n" ); document.write( "1. Tangent to the line 4x-3y=6 at (3,2) and passing through (2,-1). \n" ); document.write( "-------------- \n" ); document.write( "Find the line thru (3,2) perpendicular to the given line. \n" ); document.write( "The center will be on that line. \n" ); document.write( "---- \n" ); document.write( "Find the perpendicular bisector of the line between (3,2) and (2,-1). \n" ); document.write( "The center is also on that line. \n" ); document.write( "The intersection of the 2 lines is the center, (h,k). \n" ); document.write( "----- \n" ); document.write( "The distance from the center to either point is the radius r. \n" ); document.write( " \n" ); document.write( "-------------------------------- \n" ); document.write( "The 2 lines are \n" ); document.write( "x + 3y = 4 Eqn A \n" ); document.write( "3x + 4y = 17 Eqn B \n" ); document.write( "3x + 9y = 12 Eqn A times 3 \n" ); document.write( "------------------------------- subtract \n" ); document.write( "-5y = 5 \n" ); document.write( "y = -1 \n" ); document.write( "--- \n" ); document.write( "x = 7 \n" ); document.write( "--> center @ (7,-1) \n" ); document.write( "========================= \n" ); document.write( "Using the point (3,2): \n" ); document.write( "r^2 = diffy^2 + diffx^2 = 9 + 16 = 25 \n" ); document.write( "----- \n" ); document.write( " \n" ); document.write( "----------- \n" ); document.write( "I'll check it tomorrow. \n" ); document.write( " \n" ); document.write( " |