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Mathematics 20 Online
OpenStudy (anonymous):

LAST ONE! Find an equation of a line tangent to a circle x^2+y^2=10 at (-1. 3).

OpenStudy (anonymous):

\[2x+2yy'=0\] \[y'=-\frac{x}{y}\] plug in your point to get the slope

OpenStudy (anonymous):

\[y'=-\frac{-1}{3}=\frac{1}{3}\] is the slope, if you need the equation use the point slope formula

OpenStudy (anonymous):

y= 3x+10/3 ???

OpenStudy (anonymous):

or do i keep the 1/3 ? and make it y=1/3 x +10/3 ??

OpenStudy (anonymous):

\[y-y_1=m(x-x_1)\] with \[m=\frac{1}{3},x_1=-1,y_1=3\]

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