Find the equation of the tangent to the unit circle when x=sqrt(3)/2 (in the first quadrant)

Unit circle: x² + y² = 1 when x = sqrt(3)/2:  y² = 1 - (sqrt(3)/2)²  y² = 1 - 3/4  y² = 1/4  y = 1/2 or -1/2 (first quadrant, so y is positive, i.e. y = 1/2) find gradient at (sqrt(3)/2, 1/2):  x² + y² = 1  2x + 2y dy/dx = 0  dy/dx = -2x/2y  dy/dx = -x/y Substitute x= sqrt(3)/2, y = 1/2  dy/dx = -sqrt(3) Find equation of line:  y - y₁ = m(x - x₁)  y - 1/2 = -sqrt(3)(x-sqrt(3)/2)  y = -sqrt(3)x + 3/2 + 1/2  y = -sqrt(3)x + 2