How do you find the equations of #y = -1/3x + k# that is tangent to the unit circle in the first quadrant and find k?

Answer 1

Take the derivative of the unit circle, evaluate at #-1/3#, and use the resulting point to find #k=sqrt(10)/3#.

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#.
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Answer 2

To find the equation of a line that is tangent to the unit circle in the first quadrant, we need to determine the point of tangency. In the first quadrant, the x-coordinate will be positive, and the y-coordinate will be positive.

The equation of the unit circle is x^2 + y^2 = 1. Substituting y = -1/3x + k into this equation, we get x^2 + (-1/3x + k)^2 = 1.

Expanding and simplifying this equation, we have x^2 + (1/9)x^2 - (2/3)kx + k^2 = 1.

Combining like terms, we get (10/9)x^2 - (2/3)kx + k^2 - 1 = 0.

For the line to be tangent to the unit circle, this quadratic equation should have only one solution. This means that the discriminant (b^2 - 4ac) should be equal to zero.

Substituting the values from our equation, we have (-2/3k)^2 - 4(10/9)(k^2 - 1) = 0.

Simplifying this equation, we get 4/9k^2 - 40/9(k^2 - 1) = 0.

Expanding and simplifying further, we have 4/9k^2 - 40/9k^2 + 40/9 = 0.

Combining like terms, we get -36/9k^2 + 40/9 = 0.

Simplifying, we have 4/9k^2 = 40/9.

Dividing both sides by 4/9, we get k^2 = 10.

Taking the square root of both sides, we have k = ±√10.

Therefore, the possible values of k are √10 and -√10.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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