document.write( "Question 665773: how do i find the equation of the line tangent to the circle x^2+y^3=13 at the point (-2,3). write the answer in slope-intercept form. \n" ); document.write( "
Algebra.Com's Answer #414071 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
x^2+y^3=13 \r
\n" ); document.write( "\n" ); document.write( "The co ordinates of center are (0,0)
\n" ); document.write( "point on circle = (-2,3)\r
\n" ); document.write( "\n" ); document.write( "The slope of radius = (0-3)/(0-(-2))\r
\n" ); document.write( "\n" ); document.write( "= -3/2\r
\n" ); document.write( "\n" ); document.write( "The tangent is always perpendicular to the radius.
\n" ); document.write( "so slope of tangent will be 2/3 ( negative reciprocal)
\n" ); document.write( "the tangent passes through (-2,3)
\n" ); document.write( "m= 2/3
\n" ); document.write( "Y = m x + b
\n" ); document.write( "3.00 = 2/3 * -2 + b
\n" ); document.write( "3.00 = -1 1/3 + b
\n" ); document.write( "b= 3 + 1 1/3
\n" ); document.write( "b= 13/ 3
\n" ); document.write( "So the equation will be
\n" ); document.write( "Y = 2/3 x + 13/3
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "
\n" );