A tangent line touches a circle at exactly one point. It has a special relationship with the radius drawn to that point, and the inscribed angles of Inscribed Angles give a neat way to construct it.
💡 Why it works
A tangent is perpendicular to the radius at the point of tangency. Suppose line $\ell$ is tangent to $\odot O$ at $T$. Every point of $\ell$ other than $T$ is outside the circle, so it is farther from $O$ than $T$ is. So $T$ is the point of $\ell$ closest to $O$, and the shortest segment from a point to a line is the perpendicular one (any other segment is the hypotenuse of a right triangle, which is longer than a leg). So $\overline{OT} \perp \ell$. The converse is also true: a line perpendicular to a radius at its endpoint on the circle is tangent.
Drag $P$. The two tangent segments from $P$ are always equal, and each meets its radius at a right angle.
🔑 Key idea
A tangent is perpendicular to the radius at the point of tangency. Two tangent segments drawn to a circle from the same outside point are congruent.
✏️ Worked example: Using the right angle
$\overline{PT}$ is tangent to $\odot O$ at $T$. The radius is $8$ and $PT = 15$. Find $OP$.
$\overline{OT} \perp \overline{PT}$, so $\triangle OTP$ is a right triangle with hypotenuse $\overline{OP}$.
$OP = \sqrt{8^2 + 15^2} = \sqrt{289} = 17$.
Answer: $OP = 17$
✏️ Worked example: Two tangents from one point
$\overline{PA}$ and $\overline{PB}$ are tangent to $\odot O$. $PA = 3x + 2$ and $PB = 5x - 8$. Find $PA$.
The right triangles $OAP$ and $OBP$ share hypotenuse $\overline{OP}$ and have $OA = OB$, so they are congruent by HL, and $PA = PB$.
$3x + 2 = 5x - 8$, so $x = 5$ and $PA = 17$.
Answer: $PA = 17$
✏️ Worked example: Is it tangent?
$\odot O$ has radius $5$. Point $P$ is $13$ units from $O$, and $\overline{PA}$ reaches the circle at $A$ with $PA = 12$. Is $\overline{PA}$ tangent?
Tangent exactly when $\angle OAP$ is a right angle. Converse of the Pythagorean theorem: $5^2 + 12^2 = 169 = 13^2$.
Answer: Yes, $\overline{PA}$ is tangent at $A$.
Constructing tangents from an outside point $P$: find the midpoint $M$ of $\overline{OP}$ and draw the circle with center $M$ through $O$. It crosses $\odot O$ at two points $A$ and $B$. Since $\overline{OP}$ is a diameter of the new circle, $\angle OAP$ and $\angle OBP$ are inscribed in a semicircle: right angles. So $\overline{PA}$ and $\overline{PB}$ are tangent.
⚠️ Common mistake
Using the Pythagorean theorem with the wrong hypotenuse. The right angle is at the point of tangency, so the hypotenuse is the segment from the center to the outside point, $\overline{OP}$, not the tangent segment.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.