We see that the tangent line to the the curve #x^2+y^2=1# has a slope of #-1/3#. We can implicitly differentiate #x^2+y^2=1# to see where the slope will be equal to #-1/3#.
#d/dx[x^2+y^2=1]#
#2x+2y[dy/dx]=0#
#dy/dx=-x/y=-1/3#
We don't know any specific variables, so let's try to redefine #y# in terms of #x# using the original equation. (Only taking the positive square root because we know the answer we want is in the first quadrant.)
#y^2=1-x^2#
#y=sqrt(1-x^2)#
So, we now know:
#-1/3=-x/sqrt(1-x^2)#
#sqrt(1-x^2)=3x#
#1-x^2=9x^2#
#x=sqrt(1/10)#
If #x=1/sqrt(10)#, then #y=3/sqrt10#.
Plug this back into the original #y=mx+b# equation.
#3/sqrt10=-1/3(1/sqrt10)+b#
#10/(3sqrt10)=b#
If you want a rational denominator (or a cleaner answer): #b=sqrt10/3#
#y=-1/3x+sqrt10/3#
