Every point a fixed distance from the centre — turned into an equation.
A circle is the simplest curve to say and one of the richest to compute. Say it in words: all the points a fixed distance r from a fixed centre (a, b). That single sentence is already an equation in disguise — because in 30.1 we learned that distance is just a square root of squared differences. Square both sides to clear the root and the circle becomes (x−a)² + (y−b)² = r², an equation you can test point by point. Once a shape is an equation, every geometric question — does this line miss the circle, graze it, or cut it? do these two circles overlap? — turns into arithmetic you can finish on paper, by comparing one distance against the radius.
Forget the formula for a moment and picture the construction every child knows. Drive a pin into the page at a point C — the centre. Tie a string to it, hold a pencil at the free end, pull the string taut to some fixed length r — the radius — and walk the pencil all the way around. The taut string never changes length, so the pencil always stays the same distance from C. The mark it leaves is the circle.
That is the entire definition, and it is worth stating cleanly because the equation falls straight out of it:
A circle is the set of all points P(x, y) whose distance from a fixed centre C(a, b) equals a fixed positive radius r:
|CP| = r for every point P on the circle.
Notice what is not on the list: there is no mention of "round," no appeal to π, no formula yet. There is only a centre, a radius, and the rule |CP| = r. Everything else — the equation, the line tests, the two-circle cases — is squeezed out of that one rule by the distance formula. Let us squeeze.
Take a point P(x, y) and ask the only question that matters: is it the right distance from the centre C(a, b)? Walk the horizontal leg x − a and the vertical leg y − b; the straight span from C to P is the hypotenuse, and the Pythagorean theorem (Stage 15) — packaged as the distance formula — gives
|CP| = √( (x − a)2 + (y − b)2 ) .
The point is on the circle exactly when this equals r. Set them equal and square both sides to be rid of the awkward square root — squaring is reversible here because both sides are lengths, never negative:
(x − a)2 + (y − b)2 = r2
centre (a, b), radius r. It is nothing but "distance from the centre = r," with the root cleared away.
Read it both directions. Going forward: given a centre and a radius, you can write the equation instantly. Centre (1, 2), radius 3 gives (x − 1)2 + (y − 2)2 = 9 — note the 9 = r2, not r. Going backward: given the equation, peel off the centre (flip the signs inside the brackets) and the radius (take the square root of the right side). The equation (x + 4)2 + (y − 1)2 = 25 has centre (−4, 1) — because x + 4 = x − (−4) — and radius √25 = 5.
When the centre is the origin the equation collapses to its prettiest case, x2 + y2 = r2 — the very statement of the Pythagorean theorem for the leg lengths x and y.
The signs inside the brackets are opposite to the centre's coordinates: (a, b) sits as (x − a) and (y − b). A "+" inside means a negative coordinate. And the right side is r2, not r — if you see "= 16," the radius is 4.
The widget below lets you slide a fixed circle past a moving line; before that, dial the centre and radius yourself and watch the standard equation rewrite itself, perfectly round under equal axis scaling.
Standard form is friendly, but circles do not always arrive that way. Multiply out (x − 1)2 + (y − 2)2 = 9 and the structure scatters:
x2 − 2x + 1 + y2 − 4y + 4 = 9 ⟹ x2 + y2 − 2x − 4y − 4 = 0.
The centre and radius are still in there — just disguised. Any equation of the shape
x2 + y2 + Dx + Ey + F = 0
is the general form of a circle. To recover the centre and radius you run the disguise backward with completing the square (Stage 11): gather the x's, gather the y's, and add the magic constant to each group that turns it into a perfect square.
Take x2 + y2 − 6x + 4y − 12 = 0. Group and complete each square:
(x2 − 6x + 9) + (y2 + 4y + 4) = 12 + 9 + 4
(x − 3)2 + (y + 2)2 = 25
Half of −6 is −3 (square it: +9); half of +4 is +2 (square it: +4) — and whatever you add on the left you add on the right. The result reads off cleanly: centre (3, −2), radius √25 = 5.
You do not have to complete the square every time. Matching the worked example to the pattern gives a formula straight from the coefficients — exactly what GX.circleGen(D, E, F) computes:
For x2 + y2 + Dx + Ey + F = 0:
centre (−D2, −E2), radius r = 12√(D2 + E2 − 4F) , so r2 = D2 + E2 − 4F4.
There is one honesty check. Completing the square forces the right side to be r2 = (D² + E² − 4F)/4 — and a radius squared cannot be negative. So not every equation that looks like a circle is one.
After completing the square the right side must be positive. If D2 + E2 − 4F < 0 there is no real circle (an impossible "negative r²"); if it is exactly 0 the "circle" shrinks to a single point, the centre. For example x2 + y2 + 9 = 0 needs x² + y² = −9 — nothing in the real plane satisfies it.
Now the equations start earning their keep. Drop a straight line across the plane near a circle. There are exactly three things it can do: miss entirely, just touch at one point, or slice clean through at two points. Which one happens? You could try to solve the line and circle equations together and count solutions — and that works — but there is a faster, more visual test, and it is the heart of this whole stage.
The shortest distance from the centre to the line is the perpendicular distance d = |Ax₀ + By₀ + C| / √(A² + B²) (the formula from 30.3). Compare that single number against the radius:
| compare d and r | the line is… | how many crossings |
|---|---|---|
| d > r | apart | 0 points |
| d = r | tangent | 1 point |
| d < r | secant (crossing) | 2 points |
The reasoning is the construction itself: the closest the line ever gets to the centre is d, so if even that closest approach overshoots the radius (d > r), the whole line stays outside. If the closest approach lands exactly on the rim (d = r), the line touches at that one foot of the perpendicular — that is a tangent, and the radius to the touch point is perpendicular to the line. If the closest approach is inside (d < r), the line ducks into the disk and must come out the other side, cutting at two points.
When the line does cut, the chord it carves has a length you can read off the same right triangle. The perpendicular from the centre to a chord bisects it, so half the chord, the distance d, and the radius r form a right triangle with r as the hypotenuse:
(half-chord)2 + d2 = r2 ⟹ half-chord = √(r2 − d2), so the full chord is 2√(r2 − d2).
Circle x2 + y2 = 9 (centre (0, 0), radius 3), line 3x + 4y − 10 = 0. The centre-to-line distance is
d = |3·0 + 4·0 − 10|√(32 + 42) = 105 = 2.
Since 2 < 3 = r, the line is a secant — it cuts at two points. The chord has length 2√(32 − 22) = 2√5 ≈ 4.47.
Slide the line below and watch d cross over r. The figure draws the perpendicular drop honestly (it really lands square on the line) and marks the green crossing points wherever the engine computes them — nothing is eyeballed.
The very same idea — compare a distance against radii — sorts two circles, only now the distance is between the two centres and we compare it against the sum and the difference of the radii. Let the centres be d apart, with radii r₁ and r₂. Picture sliding one circle toward the other from far away. There are exactly five milestones:
| compare d | relationship | common points |
|---|---|---|
| d > r₁ + r₂ | separate (apart) | 0 |
| d = r₁ + r₂ | externally tangent | 1 |
| |r₁ − r₂| < d < r₁ + r₂ | intersecting | 2 |
| d = |r₁ − r₂| | internally tangent | 1 |
| d < |r₁ − r₂| | one contained in the other | 0 |
The logic mirrors the triangle inequality. Two circles share a point only when there is room for that point to sit on both rims at once. Far apart, the rims can never reach each other (d > r₁ + r₂). Just touching from outside, they kiss at one point (d = r₁ + r₂). Closer still, the rims overlap and cross twice. Slide the small one all the way inside until its rim touches the big one's rim from within (d = |r₁ − r₂|) — one touch — and closer than that the small circle floats free inside the big one, no common point at all.
Circle 1: centre (0, 0), r₁ = 2. Circle 2: centre (3, 0), r₂ = 2. The centres are d = 3 apart. Compare: r₁ + r₂ = 4 and |r₁ − r₂| = 0. Since 0 < 3 < 4, that is |r₁ − r₂| < d < r₁ + r₂ — the circles intersect at two points.
Slide the second circle along the x-axis and watch the case-name change exactly when d hits each milestone. The readout reports d, the sum, and the difference, so you can see why the engine names each case.
One last gift of the right-triangle picture. Stand at a point P outside a circle and reach two straight lines to the rim that just graze it — the two tangent lines. Each touches at one point, and at the touch point the radius is perpendicular to the tangent (the d = r rule again, applied to the tangent line). That makes a right triangle with the centre, the touch point, and P: the radius r is one leg, the tangent segment is the other leg, and the centre-to-P distance d is the hypotenuse. Pythagoras gives the tangent length at once:
From an external point P at distance d from the centre, each tangent segment has length
tangent length = √(d2 − r2) .
Both tangents from P are the same length — the triangle is the mirror image on each side. (If P is inside the circle, d < r and there is no tangent: the formula would ask for the root of a negative.)
From P(5, 0) to the circle x2 + y2 = 9 (centre (0, 0), radius 3): the distance is d = 5, so each tangent has length √(52 − 32) = √16 = 4. The two touch points sit symmetrically above and below the x-axis, each 4 from P.
One curve, told two ways, with two distance comparisons doing all the deciding.
| idea | the formula | what it tells you |
|---|---|---|
| standard form | (x − a)2 + (y − b)2 = r2 | centre (a, b), radius r directly |
| general form | x2 + y2 + Dx + Ey + F = 0 | centre (−D/2, −E/2), r² = (D²+E²−4F)/4 > 0 |
| line vs circle | compare d with r | d>r apart · d=r tangent · d<r secant |
| chord length | 2√(r2 − d2) | the full chord at centre-distance d |
| two circles | compare d with r₁±r₂ | five cases: separate → tangent → cross → tangent → contained |
| tangent length | √(d2 − r2) | each tangent from an external point P |
Next we keep one focus and add a second, then ask not for a constant distance but a constant sum of distances — and the circle stretches into an ellipse.
Plug into (x − a)² + (y − b)² = r² with a = −2, b = 3, r = 4. Since x − (−2) = x + 2: (x + 2)2 + (y − 3)2 = 16. (Right side is r² = 16, not 4.)
Complete the square: (x² − 8x + 16) + (y² + 6y + 9) = −9 + 16 + 9, i.e. (x − 4)² + (y + 3)² = 16. Centre (4, −3), radius √16 = 4. (By formula: −D/2 = 4, −E/2 = −3, r² = (64 + 36 − 36)/4 = 64/4 = 16.)
Need r² = (D² + E² − 4F)/4 > 0 with D = −2, E = 4, F = k: (4 + 16 − 4k)/4 > 0 ⟹ 20 − 4k > 0 ⟹ k < 5. At k = 5 it collapses to the single point (1, −2); for k > 5 there is no real circle.
Centre (0, 0), r = 2. Distance d = |0 + 0 − 1|/√(1² + 1²) = 1/√2 ≈ 0.707. Since 0.707 < 2, the line is a secant. Chord = 2√(r² − d²) = 2√(4 − ½) = 2√(3.5) = √14 ≈ 3.74.
d = 6, r₁ + r₂ = 5, |r₁ − r₂| = 1. Since d = 6 > 5 = r₁ + r₂, the circles are separate (apart) — no common points. (If we slid centre 2 to (5, 0), then d = 5 = r₁ + r₂ and they would be externally tangent.)
Centre (2, 0), r = 3. Distance from P to the centre: d = |8 − 2| = 6. Tangent length = √(d² − r²) = √(36 − 9) = √27 = 3√3 ≈ 5.20. Both tangents from P share this length.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the circle as a locus and derives its equation, addressing HSG-GPE.A.1 (derive the equation of a circle of given centre and radius using the Pythagorean theorem; complete the square to find the centre and radius from the general second-degree equation). The line-versus-circle and two-circle analyses, and the tangent and chord computations, draw on HSA-REI.C.7 (solve a simple system of a linear and a quadratic equation) and HSG-GPE.B.4 (use coordinates to prove geometric theorems algebraically). It builds on completing the square (Stage 11), the Pythagorean and distance ideas (Stages 15 and 30.1), and the point-to-line distance formula (30.3). A good question to ask at home: "Without graphing, how can a single number — the distance from the centre to the line — tell you whether the line misses, touches, or cuts the circle?"