How is the gradient of a tangent related to the gradient of the radius at the point of tangency?

The gradient of the tangent is the negative reciprocal of the gradient of the radius. This is because the tangent and radius are perpendicular to each other at the point of contact.

What is the difference between a tangent and a secant line?

A tangent line touches the circle at exactly one point and stays outside the circle's interior. A secant line passes through the circle, intersecting it at two distinct points.

When calculating the tangent equation, why can't you use the center of the circle as the point $(x_1, y_1)$ in the line formula?

The tangent line does not pass through the center of the circle; it only passes through the point of tangency on the circumference. Using the center would result in the equation of the radius line instead.

What happens if you forget to negate the reciprocal when finding the tangent gradient?

The resulting line will not be perpendicular to the radius. Instead, it would be a line with a related slope that likely cuts through the circle rather than just touching it.

If a radius has a gradient of $0$ (horizontal), what is the gradient of the tangent?

The gradient of the tangent would be undefined (vertical). This occurs at the 'top' or 'bottom' of the circle where the tangent line is perfectly vertical.

Define the 'Point of Tangency'.

The point of tangency is the unique coordinate where a tangent line meets the circumference of a circle. At this specific point, the line and the circle share the same gradient.

What is the point-slope form of a linear equation, and how is it used here?

The formula is $y - y_1 = m(x - x_1)$. In this context, $m$ is the negative reciprocal of the radius gradient, and $(x_1, y_1)$ is the point of tangency.

How can you use the discriminant to prove a line is a tangent?

By substituting the line equation into the circle equation, you create a quadratic. If the discriminant $b^2 - 4ac = 0$, there is only one intersection point, proving the line is a tangent.

If a point $(a, b)$ is on a circle centered at the origin, what is the gradient of the radius?

The gradient of the radius is $\frac{b}{a}$, calculated using the rise over run from the origin $(0,0)$ to the point $(a, b)$.

Why is the tangent gradient $-\frac{x}{y}$ for a circle centered at the origin?

The radius gradient is $\frac{y}{x}$. Since the tangent is perpendicular to the radius, its gradient must be the negative reciprocal, which is $-1 \div (\frac{y}{x}) = -\frac{x}{y}$.

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Equation of a Tangent

Summary

The equation of a tangent to a circle describes a straight line that touches the circumference at exactly one point. This line is fundamentally defined by its perpendicular relationship to the radius at the point of contact, allowing the use of coordinate geometry and gradient properties to derive its linear equation.

1. Definition & Core Concepts

A tangent is a straight line that touches a circle at a single specific point, known as the point of tangency. Unlike a secant line, it does not cross into the interior of the circle.
The most critical geometric property is that the tangent at any point $P$ is always perpendicular to the radius that connects the center of the circle to that point $P$ .
Because the tangent is a straight line, its equation can always be expressed in the form $y = mx + c$ or $ax + by + c = 0$ , where $m$ represents the gradient of the line.

Diagram showing a circle with a radius (green dashed line) meeting a tangent (red solid line) at point P, forming a right angle (orange square).

2. Underlying Principles

3. Methods & Techniques

4. Key Distinctions

5. Exam Strategy & Tips

6. Common Pitfalls & Misconceptions