Circle as degenerate conic (e = 0): Question bank

Definition Notation used on this page (so it's self-contained)

$P$ = a moving point on the curve; $S$ = the focus (a fixed point); $ℓ$ = the directrix (a fixed line).
$P S$ = distance from $P$ to the focus $S$ ; $P M$ = perpendicular distance from $P$ to the directrix. The ratio $P S / P M = e$ defines every conic.
$(h, k)$ = the centre of a circle; $r$ = its radius. Standard form: $(x - h)^{2} + (y - k)^{2} = r^{2}$ .
In the general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , the letter $c$ is the constant term (centre $(- g, - f)$ , radius $g^{2} + f^{2} - c$ ).
The standard ellipse (centred at the origin, long axis along $x$ ) is $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , with $a$ the semi-major and $b$ the semi-minor axis. Its foci sit at $(\pm c_{f}, 0)$ where the focal half-distance is $c_{f} = a e$ , and $b^{2} = a^{2} (1 - e^{2})$ . (We write the focal half-distance as $c_{f}$ so it never clashes with the general-form constant $c$ above.)

Before we start, one shared picture to point back at: the family of shapes you sweep through as $e$ shrinks toward $0$ . Match each curve to the legend inside the figure — that legend, not this sentence, is the source of truth for which colour is which.

Figure — Circle as degenerate conic (e = 0)

The two black dots on each ellipse are its foci; watch them slide inward toward the green centre as $e$ decreases, until on the roundest curve they have merged at the centre. Many traps below are just "which frame of this movie are we in?"

True or false — justify

Setting

P S / P M = 0

directly gives a circle.

False.

P S / P M = 0

forces

P S = 0

, so

P

is the focus — a single point. The circle appears only as the limit

e \to 0

of an ellipse, never a raw substitution.

A circle has exactly one focus, just like a parabola.

False. As

e \to 0

the ellipse's two foci both slide onto the centre, so a circle's "focus" is its centre — and every direction is equivalent, which is what total symmetry means.

Every circle has a directrix, we just usually don't draw it.

False. An ellipse has two directrices at

x = \pm a / e

; as

e \to 0

both run to infinity, so a genuine circle has no directrix at all.

If an ellipse has

a = b

then its eccentricity is

0

True. In

b^{2} = a^{2} (1 - e^{2})

, setting

b = a

gives

a^{2} = a^{2} (1 - e^{2})

, hence

1 = 1 - e^{2}

, so

e = 0

— the shape is a circle.

Increasing the radius of a circle increases its eccentricity.

False. Eccentricity measures shape (how squashed), not size. Every circle, big or small, has

e = 0

The equation

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

always represents a real circle.

False. It's a real circle only if

g^{2} + f^{2} - c > 0

; if it equals

0

you get a point circle, and if it is negative there is no real locus (an imaginary circle).

\frac{x ^{2}}{9} + \frac{y ^{2}}{9} = 1

the two axes have "no privileged direction," so it is a circle.

True. Equal denominators mean

a = b = 3

, hence

e = 0

; it's the circle

x^{2} + y^{2} = 9

, and every diameter is equivalent.

e \to 0

, the ellipse's semi-minor axis

b

grows toward the fixed semi-major axis

a

True. With

a

held fixed,

b^{2} = a^{2} (1 - e^{2}) \to a^{2}

, so

b \to a

; the "short way" catches up to the "long way" and the oval rounds out.

A circle is a conic section (a real slice of a cone), not just an algebraic curiosity.

True. Cut the double cone with a plane perpendicular to its axis and the slice is a perfect circle — it's the horizontal-cut member of the conic family.

Spot the error

"Radius

= g^{2} + f^{2} + c

, because the general form ends in

+ c

The sign is wrong:

r = g^{2} + f^{2} - c

. Expand the standard form

(x + g)^{2} + (y + f)^{2} = r^{2}

to get

x^{2} + y^{2} + 2 g x + 2 f y + (g^{2} + f^{2} - r^{2}) = 0

; matching the constant term gives

c = g^{2} + f^{2} - r^{2}

, so

r^{2} = g^{2} + f^{2} - c

— the

c

enters with a minus.

"Since

e = P S / P M

and

e = 0

, the directrix must pass through the focus."

No — you can't reason from the broken

P S / P M = 0

picture. In the limit the directrix flees to infinity, not into the focus.

"Centre of

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

(g, f)

The centre is

(- g, - f)

. Completing the square gives

(x + g)^{2} + (y + f)^{2} = \dots

, and

x + g = 0

means

x = - g

"A circle is an ellipse with

e = 1

e = 1

is a Parabola. The circle is the other extreme,

e = 0

; the whole

0 < e < 1

interval in between is ellipses.

"For the circle

x^{2} + y^{2} = a^{2}

the semi-major axis is

a

and the semi-minor axis is something smaller."

Both semi-axes equal

a

. That equality is the roundness; there is no "smaller" axis to speak of.

"Because

g^{2} + f^{2} - c

can be negative, some circles have negative radius."

A radius can't be negative. If

g^{2} + f^{2} - c < 0

there is simply no real circle — the locus is empty (an imaginary circle), not a circle with negative radius.

"To find the centre of a circle through three points, average the three points."

Averaging gives the centroid, not the centre. The centre is the point equidistant from all three (using the Distance Formula), found where perpendicular bisectors meet.

Why questions

Why do we approach the circle through the ellipse family instead of the parabola or hyperbola?

Because the ellipse is the conic "next door" to a circle — for small

e > 0

it's a barely-squashed oval, so shrinking

e

0

is a smooth, continuous relaxation into roundness.

Why does the directrix disappearing to infinity match "

e = 0

The two directrices sit at

x = \pm a / e

; as

e \to 0

both blow up to infinity. Directrices at infinity exert no directional pull, so no direction is special — exactly the symmetry of a circle.

Why does a circle "forget which direction it stretches in"?

An ellipse remembers its long axis because

a \neq = b

. When

e = 0

a = b

, so there's no longer/shorter direction to remember — every diameter is identical.

Why is a point circle (

r = 0

) called a degenerate case?

It's the shape collapsed all the way: the locus shrinks to its own centre, a single point, when

g^{2} + f^{2} - c = 0

. It's the limit where a real curve degenerates into nothing but its centre.

Why do we square both sides of

(x - h)^{2} + (y - k)^{2} = r

without worrying about extra solutions?

Both sides are non-negative (a distance and a radius), so squaring is reversible here — it can't introduce a spurious sign flip the way squaring a signed quantity would.

Why does

e

, a single number, suffice to classify all conics?

e

encodes the one degree of freedom that matters for shape — the focus-to-directrix ratio. Fix

e

and you've fixed how squashed the curve is; the rest is just position and scale.

Edge cases

What shape is

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

when

g^{2} + f^{2} - c = 0

A point circle of radius

0

— the locus is the single point

(- g, - f)

, the centre with the curve collapsed onto it.

What is the "circle" when

g^{2} + f^{2} - c < 0

An imaginary circle: there is no real locus at all, because

r^{2} < 0

has no real solution, so nothing is drawn.

In the limit

e \to 0

for the standard ellipse centred at the origin, what single point do both foci and the centre coincide at?

The origin — which is that ellipse's centre. The two foci at

(\pm c_{f}, 0)

have

c_{f} = a e \to 0

, so they slide onto the centre.

If the same limiting circle is instead written with a shifted centre

(h, k)

(or general form, centre

(- g, - f)

), where do the merged foci sit?

At that shifted centre, not the origin. The

(\pm c_{f}, 0)

formula assumes a centred ellipse; translating the whole picture by

(h, k)

carries the merged foci to

(h, k)

too, since translation moves every feature together.

Is the trivial "circle"

x^{2} + y^{2} = 0

a genuine circle?

No — its radius is

0

, so it's the degenerate point circle consisting only of the origin.

If you literally set

e = 0

P S / P M = e

, what locus do you get and why is it wrong for a circle?

You get

P S = 0

, i.e. the single point

P = S

. It's wrong because a circle is a curve of many points; the correct route is the

e \to 0

limit, not substitution.

What happens to the ellipse identity

b^{2} = a^{2} (1 - e^{2})

at the boundary

e = 1

b^{2} \to 0

, so

b \to 0

: the ellipse has degenerated out of the ellipse family entirely — this boundary is where the parabola takes over, not a circle.

Connections

Parent: Circle as a degenerate conic — the derivation these traps stress-test.
Eccentricity — the classifier $e$ every question here leans on.
Ellipse — the family the circle is the $e \to 0$ limit of.
Parabola ( $e = 1$ ) and Hyperbola ( $e > 1$ ) — the contrast cases.
Distance Formula — behind the "equidistant centre" reasoning.
Completing the Square — turns general form into $(- g, - f)$ and $r$ .
Degenerate Conics — point circle and imaginary circle live here.
Conic Sections — the cone-slice picture the circle belongs to.