document.write( "Question 1040185: find the equation of the tangents to the circle x2 +y2=9 with slope 1 \n" ); document.write( "

Algebra.Com's Answer #655009 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
Find the derivative.
\n" ); document.write( " $\"2xdx%2B2ydy=0\"$
\n" ); document.write( " $\"2ydy=-2xdx\"$
\n" ); document.write( " $\"dy%2Fdx=-x%2Fy\"$
\n" ); document.write( "So when $\"y=-x\"$ , the derivative (which is the slope of the tangent line) is equal to 1.
\n" ); document.write( "To find the points, plug this into the original equation.
\n" ); document.write( " $\"x%5E2%2B%28-x%29%5E2=9\"$
\n" ); document.write( " $\"2x%5E2=9\"$
\n" ); document.write( " $\"x%5E2=9%2F2\"$
\n" ); document.write( " $\"x=0+%2B-+3%2Fsqrt%282%29\"$
\n" ); document.write( " $\"x=0+%2B-+%283%2F2%29sqrt%282%29\"$
\n" ); document.write( "Then using the point-slope form,
\n" ); document.write( " $\"y-%283%2F2%29sqrt%282%29=1%28x%2B%283%2F2%29sqrt%282%29%29\"$
\n" ); document.write( " $\"y=x%2B3sqrt%282%29\"$
\n" ); document.write( "and
\n" ); document.write( " $\"y%2B%283%2F2%29sqrt%282%29=1%28x-%283%2F2%29sqrt%282%29%29\"$
\n" ); document.write( " $\"y=x-3sqrt%282%29\"$
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "

. \n" ); document.write( "

\n" );