document.write( "Question 1013853: Find the equation of the circle if the circle is tangent to the line 3x+y+2=0 at (-1,1) and it passes through the point (3,5). \n" ); document.write( "
Algebra.Com's Answer #630147 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! Find the equation of the circle if the circle is tangent to the line 3x+y+2=0 at A(-1,1) and it passes through the point B(3,5). \n" ); document.write( "------------ \n" ); document.write( "The center of the circle will be on the line perpendicular to the given line thru (-1,1). \n" ); document.write( "y-1 = (1/3)*(x+1) is the line. \n" ); document.write( "------------------ \n" ); document.write( "The center is equidistant from A & B so it's on the perpendicular bisector (PB) of line AB. \n" ); document.write( "The midpoint of AB is (1,3). \n" ); document.write( "The slope of AB is 1. \n" ); document.write( "--> PB is y-3 = -1*(x-1) \n" ); document.write( "Or y = -x + 4 \n" ); document.write( "--------------------- \n" ); document.write( "The center is the intersection of y = -x + 4 and y-1 = (1/3)*(x+1) \n" ); document.write( "-x + 4 = x/3 + 4/3 \n" ); document.write( "-3x + 12 = x + 4 \n" ); document.write( "x = 2 \n" ); document.write( "y = 2 \n" ); document.write( "--> the center is (2,2) \n" ); document.write( "========================= \n" ); document.write( "The radius is the distance to A or B: \n" ); document.write( " \n" ); document.write( "--------- \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |