document.write( "Question 272074: Find the equation for the line tangent to the circle x^2+y^2=58 at the points (-3,7)\r
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document.write( "I dont even know where to begin this problem. thanks for the help. \n" );
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Algebra.Com's Answer #199074 by stanbon(75887) ![]() You can put this solution on YOUR website! Find the equation for the line tangent to the circle x^2+y^2=58 at the points \n" ); document.write( "(-3,7) \n" ); document.write( "I dont even know where to begin this problem. thanks for the help. \n" ); document.write( "--------------------------------- \n" ); document.write( "The circle is the result of graphing two function: \n" ); document.write( "y = sqrt(58-x^2) \n" ); document.write( "and \n" ); document.write( "y = -sqrt(58-x^2) \n" ); document.write( "==================== \n" ); document.write( "The 1st is in the 1st and 2nd quadrants. \n" ); document.write( "The 2nd is in the 3rd and 4th quadrants. \n" ); document.write( "--------------- \n" ); document.write( "The point (-3,7) is in the 2nd quadrant so the tangency point \n" ); document.write( "is with y = sqrt(58-x^2) \n" ); document.write( "----- \n" ); document.write( "Take the derivative of y = sqrt(58 - x^2) \n" ); document.write( "y' = (1/2)(58-x^2)^(-1/2)[-2x] \n" ); document.write( "y'= (-x)/sqrt(58-x^2) \n" ); document.write( "---- \n" ); document.write( "Find the slope when x = -3 \n" ); document.write( "y'(-3) = 3sqrt(49) = 21 \n" ); document.write( "-------------------------------- \n" ); document.write( "Find the equation of the line with m = 21 passing thru (-3,7) \n" ); document.write( "7 = 21(-3)+b \n" ); document.write( "b = 70 \n" ); document.write( "----- \n" ); document.write( "Tangent Equation: \n" ); document.write( "y = 21x + 70 \n" ); document.write( "============================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |