One key unlocks every case — compare a distance with the radius.
How does a point sit relative to a circle? A line? Another circle? It looks like three different questions, but there is a single answer key for all of them: measure a distance and compare it with the radius. Closer than r, exactly r, or farther than r — that one comparison sorts a point three ways, a line three ways, and a pair of circles five ways. Master this one habit and the whole zoo of cases becomes obvious. As a bonus, the very same idea tells us exactly when three points pin down a circle — and why a triangle has one and only one circle through its corners.
Throughout, O is the center and r the radius of a circle ⊙O. The only quantity we ever change is a distance — and the only thing we ever do is hold it up against r. Keep that picture and nothing below can surprise you.
Take any point P and measure its distance d straight to the center O. There are exactly three things that can happen, and the comparison d versus r decides which:
| where P is | the test | picture |
|---|---|---|
| inside the circle | d < r | closer to O than the rim |
| on the circle | d = r | exactly on the rim |
| outside the circle | d > r | past the rim |
The circle itself is precisely the set of points with d = r — that is its definition. Everything strictly nearer (d < r) fills the disk inside; everything strictly farther (d > r) is the outside. Take it together and the disk — rim plus inside — is exactly d ≤ r.
A circle has radius r = 6. A point P sits 4 from its center. Since d = 4 < 6 = r, the point is inside the circle. If instead P were 9 from the center, then 9 > 6 and it would be outside; and at exactly 6 it would land on the rim.
A line is wider than a point, so we must be careful which distance we measure. The right one is the perpendicular distance d from the center O to the line ℓ — the shortest way from O to the line, the one that meets it at a right angle. Hold that d up against r and, once again, three cases:
| name | the test | shared points |
|---|---|---|
| apart (misses) | d > r | 0 |
| tangent (grazes) | d = r | 1 |
| secant (cuts through) | d < r | 2 |
The headline is the number of shared points. When the line runs farther than r from the center it never reaches the rim — zero points. When it runs at exactly r it just kisses the circle at one spot — a tangent, the borderline case we devote all of lesson 18.5 to. And when it runs nearer than r it plunges in one side and out the other — a secant, cutting two points. (Where d < r, those two crossings sit at the chord's ends, an offset √(r² − d²) from the foot of the perpendicular — straight from the chord rule of 18.1.)
For a point, d is the straight distance to O. For a line, d is the perpendicular distance to O — not the distance to some random spot on the line. A line can pass within a hair of the circle at one place yet still be apart if its closest approach beats r.
Let r = 5. A line whose perpendicular distance from O is 5 has d = r, so it is tangent — one shared point. A line whose distance is 2 has d = 2 < 5, so it is a secant — two shared points, sitting ±√(5² − 2²) = ±√21 ≈ ±4.58 from the foot of the perpendicular.
Now bring two circles together — radii R and r (say R ≥ r), centers a distance D apart. The single comparison becomes a double one: weigh D against the sum R + r and the difference R − r. Two comparisons, and out fall five arrangements:
| arrangement | the test | common points |
|---|---|---|
| separate (outside each other) | D > R + r | 0 |
| externally tangent | D = R + r | 1 |
| intersecting | R − r < D < R + r | 2 |
| internally tangent | D = R − r | 1 |
| contained (one inside the other) | D < R − r | 0 |
Read it as a journey. Start with the circles far apart (separate, 0 points) and walk the second one in. The moment their rims first touch from outside, D = R + r — externally tangent, one point. Push closer and the rims overlap, crossing at two points — intersecting. Keep going until the small circle's rim touches the big one's from the inside: D = R − r — internally tangent, one point again. Push even closer and the small circle is swallowed whole — contained, 0 points. (When D = 0 the centers coincide; if the radii differ these are concentric circles.)
Two circles have radii R = 5 and r = 3, so R + r = 8 and R − r = 2. If their centers are 10 apart, then D = 10 > 8 — they are separate. If the centers are 2 apart, then D = 2 = R − r — they are internally tangent (the little one touches the big one from inside). And at D = 6, since 2 < 6 < 8, they intersect in two points.
The same idea — equal distances from a center — answers a different-looking question: how many points does it take to nail down one circle?
Through one point you can draw infinitely many circles — center it anywhere and pick a matching radius. Through two points A and B you still have infinitely many: a circle through both must have its center the same distance from each, so the center can be any point on the perpendicular bisector of AB — a whole line of choices. But pin down a third point C, not on the line through A and B, and the freedom vanishes.
Here is why. The center must be equidistant from A and B, so it lies on the perpendicular bisector of AB. It must also be equidistant from B and C, so it lies on the perpendicular bisector of BC. Two distinct lines that aren't parallel meet in exactly one point — call it the circumcenter. That single point is equidistant from all three, so it is the center of the one and only circle through A, B, and C. The matching radius is its common distance to them.
Three points not in a line determine exactly one circle. Its center is the circumcenter — where the perpendicular bisectors of the connecting segments meet — and the circle is the triangle's circumscribed circle. So every triangle has one, and only one, circle through its three corners.
What if the three points are on a line? Then the perpendicular bisectors of AB and BC are both perpendicular to that same line, so they are parallel and never meet — there is no center, and no circle passes through three collinear points.
Point, line, or another circle — every position question is the same question in disguise: find the relevant distance and compare it with r (or, for two circles, with R + r and R − r). Counting the shared points is then automatic.
One comparison runs the whole lesson — a distance held up against the radius.
| question | distance to compare | cases (and shared points) |
|---|---|---|
| point P vs circle | d = OP (straight to O) | inside d<r · on d=r · outside d>r |
| line ℓ vs circle | d = ⟂ distance from O to ℓ | apart d>r (0) · tangent d=r (1) · secant d<r (2) |
| two circles (R, r), D apart | D vs R + r and R − r | separate (0) · ext. tangent (1) · intersecting (2) · int. tangent (1) · contained (0) |
A point P is 4 from the center of a circle of radius 6. Is P inside, on, or outside the circle?
Compare d with r: 4 < 6, so P is inside the circle.
A line lies 5 from the center of a circle of radius 5 (perpendicular distance). How many points does it share with the circle, and what is the line called?
Here d = 5 = r, so the line is tangent — exactly 1 shared point.
Two circles have radii 3 and 5, and their centers are 10 apart. How do they sit relative to each other?
R + r = 5 + 3 = 8. Since D = 10 > 8, the circles are separate — outside each other, with 0 common points.
Two circles have radii 4 and 6, and their centers are 2 apart. How do they sit?
R − r = 6 − 4 = 2. Since D = 2 = R − r, the circles are internally tangent — the small one touches the big one from inside, at 1 point.
How many circles pass through three given points that do not lie on a line? Why?
Exactly one. The center must be equidistant from all three, so it lies on the perpendicular bisector of each connecting segment; two of those bisectors meet at a single point — the circumcenter — which fixes the one circle.
A line lies 2 from the center of a circle of radius 5. How many points do they share, and exactly where (relative to the foot of the perpendicular)?
d = 2 < 5 = r, so it is a secant with 2 shared points. They sit ±√(r² − d²) = ±√(25 − 4) = ±√21 ≈ ±4.58 along the line from the foot of the perpendicular.
Six questions to lock it in. Tap the answer you think is right.
The whole lesson is one move repeated three times: measure the right distance and compare it with the radius. That single habit unifies what textbooks often present as three unrelated lists of cases. Encourage students to always name the distance first ("d is the perpendicular from O to the line"; "D is the distance between the two centers") before they reach for a verdict — most errors here are not arithmetic but measuring the wrong thing.
Three traps recur. First, for a line, students compare the distance to some point on the line rather than the perpendicular distance to O. Second, in the two-circle picture they forget the fifth case — one circle fully contained inside the other (D < R − r, still 0 points), lumping it in with "intersecting." Third, they assume any three points fix a circle; collinear points do not, because the perpendicular bisectors come out parallel and never meet.
This lesson aligns with the Common Core standards G-C.A.2 (relationships between a line and a circle), G-C.A.3 (construct the inscribed and circumscribed circles of a triangle; the circumcenter), and G-CO.D.12 (formal geometric constructions). It also sets up lesson 18.5, where tangency — the borderline d = r case — gets its own properties, test, and the inscribed circle of a triangle.