Tangent to a Circle – Properties, Theorem, Examples (Geometry)

Contents

Key takeaway: A tangent to a circle is a line touching the circle at exactly one point, and it always forms a right angle with the radius drawn to that point of tangency.

Grade 10 – GeometryG-C.A.4Geometry and Measurement

Definition of Tangent to a Circle

Definition

A tangent to a circle grazes it like a ruler held against a coin edge: the two touch at a single spot, the point of tangency, and the line never slips inside.

The key fact is the square corner. A radius ending at the touching spot meets the line at \(90^{\circ}\). From any point outside you can draw exactly two such lines, and the pieces running from that outside point to the two touching spots have equal length. These facts turn many circle problems into right-triangle problems.

Formula and notation

If \(T\) is the point of tangency, \(O\) the center and \(P\) an outside point on the tangent line:

\[ \overline{OT} \perp \overleftrightarrow{PT}, \qquad PT^2 = OP^2 – r^2 \]

The second equation is the Pythagorean theorem in right triangle \(OTP\), with \(r = OT\).

Both tangent segments from P are 12 long, and each meets its radius at a right angle.

Key facts

The radius to the point of tangency forms a 90 degree angle with the tangent.
Two tangent segments from one outside point are equal in length.
A tangent line meets the circle once; a secant meets it twice.
The center lies on the line that bisects the angle between two tangents.

Tangent to a Circle example

Problem

A round skating rink has a 9 m radius. A boardwalk begins at a gate \(P\), 15 m from the rink’s center \(O\), and runs straight until it just brushes the edge at \(T\). How long is the boardwalk?

The walk is a tangent to a circle, so the radius \(OT\) is perpendicular to it at \(T\).
Triangle \(OTP\) has its right angle at \(T\), so \(PT^2 = OP^2 – OT^2 = 225 – 81 = 144\).
Take the square root: \(PT = 12\).

The boardwalk is 12 m long; one on the other side would also be 12 m.

Common mistake

Watch out

Ira treats the 15 m as the hypotenuse’s partner and adds: \(PT = \sqrt{15^2 + 9^2} \approx 17.5\) m.

Why it is wrong

The distance from the gate to the center is the hypotenuse of right triangle \(OTP\), because the right angle is at \(T\). Subtract: \(\sqrt{15^2 – 9^2} = 12\). The tangent can never be longer than the gate’s distance to the center.

Related terms and lessons

Geometry · Parabolas from Focus and Directrix, and Tangent Lines to a CircleLesson Worksheet Quiz

Browse the glossary: A–Z list · Geometry vocabulary · Geometry and Measurement

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