Implicit Differentiation: Slope on a Circle

A circle is not the graph of a function, since most x-values have two points above them, but it still has a tangent at every point and the tangent has a slope. Implicit differentiation finds it without solving for y: differentiate both sides of x² + y² = r² with respect to x, treating y as a function of x, and get 2x + 2y·dy/dx = 0, so dy/dx = −x/y. Worked example: on x² + y² = 25 at the point (3, 4), the slope is −3/4 = −0.75. At (3, −4), on the lower half, the same formula gives +0.75. The radius to (3, 4) has slope 4/3, and 4/3 × (−3/4) = −1: the tangent is perpendicular to the radius, which is the geometry the calculus is confirming.

The classic mistake is differentiating y² as 2y and forgetting the dy/dx. y is a function of x, so the chain rule applies: d/dx(y²) = 2y·dy/dx. Without that factor there is no dy/dx in the equation to solve for. At (±r, 0) the formula divides by zero, correctly: the tangent there is vertical. For a circle centred at (h, k) instead of the origin the slope is −(x − h)/(y − k). The same move handles any curve written as a relation between x and y, such as an ellipse, and it is exactly what every related-rates problem does with time t in place of x.