Find the points where the line $ y = mx + b $ is tangent to the unit circle

To find the points where the line $ y = mx + b $ is tangent to the unit circle, we start with the equation of the unit circle:

$ x^2 + y^2 = 1 $

Substitute $ y = mx + b $ into the unit circle equation:

$ x^2 + (mx + b)^2 = 1 $

Expand the equation:

$ x^2 + m^2x^2 + 2mxb + b^2 = 1 $

Combine like terms:

$ (1 + m^2)x^2 + 2mxb + b^2 = 1 $

This is a quadratic equation in $ x $:

$ (1 + m^2)x^2 + 2mxb + (b^2 – 1) = 0 $

For the line to be tangent to the circle, the discriminant must be zero:

$ (2mb)^2 – 4(1 + m^2)(b^2 – 1) = 0 $

Simplify the discriminant:

$ 4m^2b^2 – 4(1 + m^2)(b^2 – 1) = 0 $

Solving this will give the condition on $ b $.

To find the point of tangency for the line $ y = mx + b $ with the unit circle:

$ x^2 + y^2 = 1 $

Substitute $ y = mx + b $:

$ x^2 + (mx + b)^2 = 1 $

Expand and simplify:

$ (1 + m^2)x^2 + 2mxb + b^2 = 1 $

Quadratic form:

$ (1 + m^2)x^2 + 2mxb + (b^2 – 1) = 0 $

For tangency, the discriminant must be zero:

$ 4m^2b^2 – 4(1 + m^2)(b^2 – 1) = 0 $

To find the points where the line $ y = mx + b $ is tangent to the unit circle, we use:

$ x^2 + y^2 = 1 $

Substitute $ y = mx + b $:

$ x^2 + (mx + b)^2 = 1 $

Expand and simplify:

$ (1 + m^2)x^2 + 2mxb + b^2 = 1 $

Condition for tangency:

$ 4m^2b^2 – 4(1 + m^2)(b^2 – 1) = 0 $

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