The formula for a circle is:
#(x-h)^2+(y-k)^2=r^2#
Where #(h,k)# is the center of the circle and #r# is the radius.
The unit circle is a particular type of circle with center #(0,0)# (the origin) and radius #1#, so its equation is:
#x^2+y^2=1#
We are told that the line #y=-1/3x+k# (which has slope #-1/3# and #y#-intercept #k#) is tangent to this circle in the first quadrant. In order to determine #k#, we'll need the point of tangency - that is, the point #(x,y)# where the line touches the circle.
Now, since the slope of the tangent line is #-1/3# at this unknown point, we know the derivative of the unit circle is #-1/3# at that point. Why? Because the slope of the tangent line is the definition of the derivative! So in order to find the point of tangency, we'll find the derivative of the unit circle, set it equal to #-1/3#, and solve.
Taking the derivative of #x^2+y^2=1# will require implicit differentiation:
#d/dx(x^2+y^2)=d/dx(1)#
#2x+2ydy/dx=0#
#x+ydy/dx=0#
#dy/dx=-x/y#
Since #x^2+y^2=1#, solving for #y# yields #y=sqrt(1-x^2)#. That means
#dy/dx=-x/(sqrt(1-x^2))#
Now we set this equal to #-1/3#:
#-1/3=-x/(sqrt(1-x^2))#
#1/3=x/(sqrt(1-x^2))#
#1/3sqrt(1-x^2)=x#
#1/9(1-x^2)=x^2#
#1-x^2=9x^2#
#1=10x^2->x=+-sqrt(1/10)#
Since we're dealing with the first quadrant, we use the positive square root:
#x=sqrt(1/10)#
Alright, almost there! Now we plug this into #y=sqrt(1-x^2)# to solve for #y#:
#y=sqrt(1-x^2)#
#y=sqrt(1-(sqrt(1/10))^2)#
#y=sqrt(1-1/10)=sqrt(9/10)=3/sqrt(10)#
Yay! We found out point of tangency: #(sqrt(1/10),3/sqrt(10))#. Finally, we can plug this into #y=-1/3x+k# to find #k#:
#y=-1/3x+k#
#3/sqrt(10)=-1/3sqrt(1/10)+k#
#3/sqrt(10)+1/3sqrt(1/10)=k#
#3/sqrt(10)+1/(3sqrt(10))=k#
#9/(3sqrt(10))+1/(3sqrt(10))=k#
#k=(10)/(3sqrt(10))=sqrt(10)/3#
Therefore the equation of the tangent line is #y=-1/3x+sqrt(10)/3#.
