At the single point of contact the radius stands perfectly upright — and two tangents from a point are twins.
Of all the ways a line can meet a circle, the most delicate is the single touch — the tangent, the line that kisses the rim at exactly one point and then leaves. That one point of contact hides a perfect right angle: the radius drawn to it stands exactly upright on the line. Turn that fact into a test, draw two tangents from an outside point and find them equal, and you can tuck a circle snugly inside any triangle. Tangency is the borderline case from the last lesson — and the richest one in the whole stage.
Recall the three ways a line ℓ can sit against a circle, sorted by the distance d from the center O to the line: if d > r it misses; if d < r it cuts through twice (a secant); and right between them, when d = r exactly, the line just grazes the circle at a single point. That borderline line is a tangent, and the one point where it touches is the point of tangency T.
Here is the secret of that single point. Because the line only reaches the circle and never goes inside, T is the closest point of the whole line to the center — the perpendicular from O lands exactly there. And the shortest segment from a point to a line is always the perpendicular one. So the radius drawn to the touch point is perpendicular to the tangent:
OT ⊥ ℓ — the radius at the point of contact meets the tangent at a right angle.
The radius to the point of tangency is perpendicular to the tangent. Equivalently: the perpendicular distance from the center to a tangent line equals the radius r — the smallest distance a line to the circle can have.
Move T anywhere on the circle. The tangent there always meets the radius OT at exactly 90°.
A good theorem runs both ways. We just saw: if a line is tangent, then it is perpendicular to the radius at the touch point. Now read it backward. Start at a point T on the circle, draw the radius OT, and put down the line through T that is perpendicular to OT. Is that line a tangent?
Yes — and the reason is the same distance argument. The foot of the perpendicular from O to this line is T itself, so the line's distance from O is exactly |OT| = r. A line whose distance from the center equals r meets the circle at one point only. So the line is a tangent, touching at T.
A line is tangent to a circle ⇔ it is perpendicular to the radius at the point where it meets the circle.
→ Property: tangent ⇒ ⊥ to the radius (use it to get a right angle). ← Test: ⊥ to the radius ⇒ tangent (use it to build a tangent).
A tangent does not cross into the circle — it only touches the edge. If a line ever dips inside, it has crossed the rim twice and is a secant, not a tangent. "One point, and the radius perpendicular" is the whole signature of tangency.
Stand at a point P outside the circle. From there you can draw two tangent lines — one swinging up to touch at A, one down to touch at B. (From a point on the circle there is exactly one tangent; from a point inside, none.) The beautiful fact is that those two tangent segments come out equal:
PA = PB, and PO bisects both ∠APB and ∠AOB.
Here is the reasoning, said out loud. Look at the two triangles △OAP and △OBP. Each has a right angle — at A and at B — because the radius is perpendicular to its tangent (18.5.1). The two radii are equal, OA = OB = r. And they share the same hypotenuse, OP. A right angle, an equal leg, and a shared hypotenuse: that is exactly the HL congruence test. So △OAP ≅ △OBP, which forces the remaining parts to match — PA = PB, and the angles at P (and at O) split evenly, so PO is the bisector.
And we can read off the common length with the Pythagorean theorem. In right triangle △OAP the hypotenuse is OP, one leg is the radius r, and the other leg is the tangent PA:
PA = PB = √(OP² − r²).
A point P sits OP = 13 from the center of a circle of radius r = 5. How long is each tangent from P?
PA = PB = √(OP² − r²) = √(13² − 5²) = √(169 − 25) = √144 = 12. The two tangents are 12 long apiece — a clean 5–12–13 right triangle hiding in the figure.
Slide P along the ray from O. The two tangents stay equal — and their length is always √(OP² − r²).
Now put tangents to work. Take any triangle and tuck the largest possible circle inside it, just big enough to touch all three sides at once. That is the triangle's inscribed circle, or incircle. Each side is a tangent to it, so by 18.5.1 the radius drawn to each touch point is perpendicular to that side.
Where is its center? The center must be the same distance from all three sides — and the set of points equidistant from two sides of an angle is exactly that angle's bisector. A point equidistant from all three sides therefore lies on all three angle bisectors at once. So the three angle bisectors of a triangle meet at a single point — the incenter I — and that point is the center of the incircle. Its common distance to the three sides is the inradius.
Because the incenter is built from the interior angle bisectors, it always lands inside the triangle — every triangle, no exceptions.
The incenter is the meeting of the three angle bisectors; it is equidistant from the three sides, and that distance is the inradius. The incircle touches each side at the foot of the perpendicular from I. The incenter is always inside the triangle.
Handy formulas: with side lengths a, b, c, semiperimeter s = (a+b+c)/2, and area K, the inradius is r = K / s.
For a 6–8–10 right triangle: the area is K = ½·6·8 = 24 and the semiperimeter is s = (6+8+10)/2 = 12. So the inradius is r = K/s = 24/12 = 2. The incircle of radius 2 sits 2 units in from each leg of the right angle.
A triangle carries two famous centers, and they are easy to mix up. The incenter we just met. Its cousin, the circumcenter from 18.4, is the center of the circumscribed circle that passes through all three corners. They are built from different lines and answer different questions:
| Incenter | Circumcenter | |
|---|---|---|
| built from | angle bisectors | perpendicular bisectors |
| equidistant from | the three sides | the three vertices |
| its circle | inscribed (touches sides) | circumscribed (through corners) |
| radius | inradius = K / s | circumradius = abc / (4K) |
| position | always inside | inside, on, or outside |
For our 6–8–10 right triangle, the circumcenter lands at the midpoint of the hypotenuse, so the circumradius is half the hypotenuse: 10 / 2 = 5 (this is the "angle in a semicircle" fact from 18.3, read in reverse). Compare with the inradius of 2 — same triangle, two very different circles.
Switch between the inscribed circle, the circumscribed circle, and both. Watch where each center sits and what its circle touches.
One delicate touch, and a chain of consequences:
A line is tangent to a circle at T, and O is the center. What is the angle between the radius OT and the tangent line?
90°. The radius to the point of tangency is always perpendicular to the tangent (tangent property).
From a point P that is 13 away from the center of a circle of radius 5, a tangent is drawn. How long is the tangent segment from P to the point of contact?
Tangent length = √(OP² − r²) = √(13² − 5²) = √(169 − 25) = √144 = 12. (A 5–12–13 right triangle.)
Two tangents are drawn to a circle from one external point, touching at A and B. What can you say about the lengths PA and PB, and why?
They are equal: PA = PB. The triangles △OAP and △OBP are congruent by HL (right angles at A and B, equal radii OA = OB, shared hypotenuse OP), so the matching tangent sides are equal.
The inscribed circle of a triangle has its center at the meeting point of which three lines?
The three angle bisectors. Their common point — the incenter — is equidistant from all three sides, which is exactly the inradius.
For a triangle, the incenter is always inside it. But the circumcenter can land where? Give the case.
Outside the triangle — this happens for an obtuse triangle. (For a right triangle the circumcenter sits exactly on the hypotenuse's midpoint; for an acute triangle it is inside.)
A right triangle has legs 6 and 8 and hypotenuse 10. Find its inradius and its circumradius.
Area K = ½·6·8 = 24, semiperimeter s = (6+8+10)/2 = 12, so inradius = K/s = 24/12 = 2. The circumcenter is the hypotenuse's midpoint, so circumradius = 10/2 = 5.
Six questions to lock it in. Tap the answer you think is right.
This lesson turns the borderline case of line-and-circle positions into its own rich theory. The single big idea is the tangent property: the radius to the point of contact is perpendicular to the tangent. Everything else follows from it — the tangent test (its converse), the equal-tangent theorem (proved by HL congruence), and the inscribed circle of a triangle. Encourage students to reason out loud: name the right angle, name the equal radii, name HL, then read off the conclusion.
The misconception to watch is twofold. First, students often imagine a tangent crossing into the circle — stress that it only touches the edge at one point, never dips inside. Second, they routinely swap the incenter and the circumcenter. Keep them straight by their construction: incenter from angle bisectors, equidistant from the sides, always inside; circumcenter from perpendicular bisectors, equidistant from the vertices, possibly outside. The side-by-side table and the toggle widget are built to drill exactly this distinction.
Common Core: this lesson supports G-C.A.2 (the radius is perpendicular to a tangent), G-C.A.3 (the inscribed circle of a triangle; the incenter as the meeting of the angle bisectors), and G-C.A.4 (constructing a tangent line to a circle from a point outside it — the tangent test makes this possible).