# 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?

Take the derivative of the unit circle, evaluate at

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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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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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