Circle as degenerate conic (e = 0): Worked examples

Before anything else, the symbols we will lean on constantly:

$(h, k)$ = the centre coordinates (horizontal $h$ , vertical $k$ ).
$r$ = the radius, the fixed distance from centre to every point on the circle.
General form $x^{2} + y^{2} + 2 g x + 2 f y + c$ , from which centre $= (- g, - f)$ and $r = g^{2} + f^{2} - c$ . Here $g, f$ are just half the coefficients of $x, y$ , and the letter $c$ is the constant term of the circle equation — nothing else on this page.
$e$ = the eccentricity, a single number measuring how squashed a conic is; $e = 0$ is perfectly round (circle). We only use $e$ inside the ellipse-limit example, and we will re-explain it there.

If any of those still feel loose, the parent note built them from the Distance Formula first.

The scenario matrix

Every circle problem you can be handed falls into one of these cells. The right-hand column names the example that clears it.

Cell	What makes it tricky	Cleared by
A. Centre in each quadrant	signs of $h, k$ (so signs of $g, f$ ) flip	Ex 1
B. Real circle	$g^{2} + f^{2} - c > 0$ : ordinary radius	Ex 1, 2
C. Point circle	$g^{2} + f^{2} - c = 0$ : radius $0$ , a single point	Ex 3
D. Imaginary circle	$g^{2} + f^{2} - c < 0$ : no real points at all	Ex 4
E. Ellipse → circle limit	$e \to 0$ , $a = b$	Ex 5
F. Circle → focus limit	naive $e = 0$ collapse to a point	Ex 6
G. Circle through 3 points	equidistance ( $e = 0$ symmetry)	Ex 7
H. Centre ON an axis	$h = 0$ or $k = 0$ (edge case, tangent word problem)	Ex 8
I. Exam twist	scaled equation $2 x^{2} + 2 y^{2} + \dots$	Ex 9

We now hit each cell in order. The first figure below reads left to right: a coral dot marks the imaginary regime ( $g^{2} + f^{2} - c < 0$ , no real points), a butter dot the point-circle ( $= 0$ ), and the mint band the family of ordinary real circles ( $> 0$ ). Keep that colour code in mind as the examples march through each region.

The worked examples

Worked example Example 1 — Cell A & B: centre in every quadrant

For each equation, find the centre and radius: (i) $x^{2} + y^{2} - 6 x - 8 y + 9 = 0$ , (ii) $x^{2} + y^{2} + 6 x - 8 y + 9 = 0$ , (iii) $x^{2} + y^{2} + 6 x + 8 y + 9 = 0$ , (iv) $x^{2} + y^{2} - 6 x + 8 y + 9 = 0$ .

Forecast: the number $9$ never changes and $∣6∣, ∣8∣$ never change, so all four have the same radius — only the centre jumps between quadrants. Guess the radius before reading on.

Read $g, f, c$ off the general form. For (i): $2 g = - 6 \Rightarrow g = - 3$ ; $2 f = - 8 \Rightarrow f = - 4$ ; $c = 9$ . Why this step? The general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ stores the centre in $g, f$ directly.
Centre $= (- g, - f)$ . (i) gives $(3, 4)$ — quadrant I (both positive). Why this step? Completing the square turns $x^{2} + 2 g x$ into $(x + g)^{2}$ , whose centre-coordinate is $- g$ .
Flip the signs for the others. (ii) $g = 3, f = - 4 \Rightarrow (- 3, 4)$ quadrant II; (iii) $g = 3, f = 4 \Rightarrow (- 3, - 4)$ quadrant III; (iv) $g = - 3, f = 4 \Rightarrow (3, - 4)$ quadrant IV. Why this step? The centre sign is opposite the coefficient sign — this is the single most common slip.
Radius (same for all). $r = g^{2} + f^{2} - c = 9 + 16 - 9 = 16 = 4$ . Why this step? $r^{2} = g^{2} + f^{2} - c$ ; here it is $9 + 16 - 9 = 16$ .

Verify: plug the centre of (i) into $(x - 3)^{2} + (y - 4)^{2} = 16$ — expand: $x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = 16$ , i.e. $x^{2} + y^{2} - 6 x - 8 y + 9 = 0$ . ✓ Matches. Each quadrant is covered.

The next figure draws all four of these circles at once. Notice each is the same size (radius $4$ ) but its centre sits in a different quadrant — the coloured curve for quadrant I is lavender, II coral, III mint, IV butter. The picture makes the "sign flip only moves the centre" claim visible.

Worked example Example 5 — Cell E: an

Ellipse relaxing into a circle An ellipse has semi-major axis $a = 10$ . Find $b$ at eccentricity $e = 0.8$ , then describe $e \to 0$ .

Recall $e$ (eccentricity) measures squash: $e = 0$ round, close to $1$ very flat. Here $a$ is the long half-width and $b$ the short half-width. Note: in this example we will also meet the ellipse's focal distance, which we deliberately call $d$ (NOT $c$ , to avoid clashing with the circle's constant term $c$ from every other example).

Forecast: big $e$ → very flat; as $e \to 0$ the two axes should meet at $b = a = 10$ .

Use $b^{2} = a^{2} (1 - e^{2})$ (the Eccentricity identity). At $e = 0.8$ : $b^{2} = 100 (1 - 0.64) = 36$ , so $b = 6$ . Why this step? This is the tool that links squash-factor $e$ to the short axis $b$ .
Send $e \to 0$ : $b^{2} \to 100 (1 - 0) = 100$ , so $b \to 10 = a$ . Why this step? Equal axes are the signature of a circle.
The equation $\frac{x ^{2}}{100} + \frac{y ^{2}}{100} = 1$ becomes $x^{2} + y^{2} = 100$ , radius $10$ . Why this step? With $a = b = 10$ both denominators are $100$ ; multiplying through by $100$ turns the ellipse equation into the circle equation, showing the ellipse literally becomes a circle at $e = 0$ .

Verify: the focal distance is $d = a e$ ; at $e = 0.8$ , $d = 8$ , so foci sit at $\pm 8$ ; as $e \to 0$ , $d \to 0$ , both foci hug the centre. ✓

Worked example Example 6 — Cell F: why the naive

$e = 0$ substitution collapses A student writes $\frac{P S}{P M} = 0 \Rightarrow P S = 0$ . What shape is that, really?

Forecast: guess whether $P S = 0$ is a circle or something smaller.

$P S = 0$ means point $P$ has zero distance to focus $S$ , so $P = S$ . Why this step? Only one point is at distance $0$ from $S$ .
That is a single point, not a circle — the degenerate limit pointed the wrong way. Why this step? The correct route is to shrink $e$ through the ellipse (Ex 5), keeping $a$ fixed, so the shape stays finite.
Correct statement: the circle is $lim_{e \to 0} (ellipse of fixed a)$ , not the substitution $e = 0$ into $P S / P M$ . Why this step? This pins down the one legitimate meaning of " $e = 0$ gives a circle": it is a limit of finite ellipses (radius stays $a$ ), not an algebraic substitution (which wrongly forces radius $0$ ).

Verify: with fixed $a = 10$ , $r = a = 10 \neq = 0$ ; the naive route gives $r = 0$ . The two disagree, proving the naive substitution is invalid. ✓

Worked example Example 7 — Cell G: circle through three points (the

$e = 0$ symmetry) Find the circle through $A (1, 1)$ , $B (5, 1)$ , $C (1, 5)$ .

Forecast: the centre is equidistant from all three (no preferred direction — that's $e = 0$ ). $A, B$ share a $y$ ; $A, C$ share an $x$ ; expect a tidy centre.

Equidistant from $A, B$ : $(h - 1)^{2} = (h - 5)^{2} \Rightarrow - 2 h + 1 = - 10 h + 25 \Rightarrow 8 h = 24 \Rightarrow h = 3$ . Why this step? The $y$ -parts cancel, isolating $h$ .
Equidistant from $A, C$ : $(k - 1)^{2} = (k - 5)^{2} \Rightarrow k = 3$ by the same algebra. Why this step? $A$ and $C$ share the same $x$ , so now the $x$ -parts cancel and $k$ is isolated the same way $h$ was — one equation per unknown.
Centre $(3, 3)$ ; radius $r = (3 - 1)^{2} + (3 - 1)^{2} = 4 + 4 = 8 = 22$ . Why this step? Distance from centre to any of the three points must match.
Equation: $(x - 3)^{2} + (y - 3)^{2} = 8$ . Why this step? We plug the found centre $(3, 3)$ and $r^{2} = 8$ into the standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ to write the answer as one clean equation.

Verify: check $C (1, 5)$ : $(1 - 3)^{2} + (5 - 3)^{2} = 4 + 4 = 8$ . ✓ All three lie on it.

The figure below shows this circle with the three points on it and the three equal mint dashes from the centre to $A, B, C$ — equal length is exactly the $e = 0$ "no preferred direction" symmetry made visible.

Worked example Example 8 — Cell H: centre ON an axis (word problem, tangent condition)

A garden sprinkler sits at $(3, 4)$ (metres) and its spray just reaches the straight wall along the $x$ -axis ( $y = 0$ ). Find the watered region's boundary equation and its area. Then note what would happen if the sprinkler were at $(0, 4)$ (centre on the $y$ -axis).

Forecast: "just reaches the wall" means the circle is tangent to the $x$ -axis, so the radius equals the height of the centre above it — guess $r = 4$ .

Centre $(3, 4)$ , and tangent to $y = 0$ means $r =$ perpendicular distance from centre to that line $= ∣4∣ = 4$ . Why this step? Tangency = the closest point of the line is exactly $r$ away; that distance is just the $y$ -coordinate here.
Boundary: $(x - 3)^{2} + (y - 4)^{2} = 16$ . Why this step? We insert centre $(3, 4)$ and $r^{2} = 16$ into the standard form — the boundary of the watered disk is the circle of that radius about the sprinkler.
Area $= π r^{2} = 16 π \approx 50.27 m^{2}$ . Why this step? Units check: $r$ in metres, so $r^{2}$ in $m^{2}$ , correct for area.
Edge case $h = 0$ : if the sprinkler were at $(0, 4)$ , its centre lies on the $y$ -axis; the equation is $x^{2} + (y - 4)^{2} = 16$ (no $x$ -shift). It is still tangent to the $x$ -axis, and it is now symmetric about the $y$ -axis. Why this step? When a centre coordinate is $0$ the corresponding shift term vanishes — a genuinely different-looking equation that beginners misread as "missing a term."

Verify: the lowest point of the spray is $(3, 4 - 4) = (3, 0)$ , which lies on the wall $y = 0$ — tangent, exactly one touch. ✓

Worked example Example 9 — Cell I: exam twist (a scaled equation)

Find centre and radius of $2 x^{2} + 2 y^{2} - 8 x + 12 y - 6 = 0$ .

Forecast: the general-form rules need the $x^{2}$ coefficient to be $1$ . Divide first, or the numbers lie to you.

Divide through by $2$ : $x^{2} + y^{2} - 4 x + 6 y - 3 = 0$ . Why this step? $g, f, c$ are only read off correctly when the leading coefficients are $1$ (and equal). Forgetting this is the classic trap.
$g = - 2$ , $f = 3$ , $c = - 3$ ; centre $= (2, - 3)$ . Why this step? Now that the equation is in true general form, we halve the linear coefficients for $g, f$ and negate for the centre — the same reading-off used since Ex 1.
$r = g^{2} + f^{2} - c = 4 + 9 - (- 3) = 16 = 4$ . Why this step? Apply the radius formula $r = g^{2} + f^{2} - c$ to the divided coefficients; the $- c = + 3$ is the sign-care point again.

Verify: complete the square on the divided equation: $(x - 2)^{2} - 4 + (y + 3)^{2} - 9 - 3 = 0 \Rightarrow (x - 2)^{2} + (y + 3)^{2} = 16$ , radius $4$ , centre $(2, - 3)$ . ✓ If you had not divided, you would wrongly read $g = - 4$ and get a bogus radius.

Recall check

Below, each line is a question followed by its answer, separated by the three-colon marker ::: (Obsidian hides the answer until you reveal it — treat everything left of ::: as the prompt, everything right as the hidden answer).

Recall What guarantees a circle is real, a point, or imaginary?

Sign of $g^{2} + f^{2} - c$ ::: $> 0$ real circle, $= 0$ point circle, $< 0$ imaginary (no real points). First move on $2 x^{2} + 2 y^{2} + \dots = 0$ ? ::: Divide by $2$ so the $x^{2}$ coefficient is $1$ before reading $g, f, c$ . "Circle tangent to the $x$ -axis, centre $(h, k)$ " gives what radius? ::: $r = ∣ k ∣$ . Centre on the $y$ -axis means which coordinate is zero? ::: $h = 0$ , so the $x$ -shift term disappears.

Connections

Hinglish version — same content in Hinglish.
Conic Sections · Eccentricity · Ellipse · Parabola · Hyperbola — the family this circle belongs to.
Distance Formula — builds the equidistance conditions (Ex 7).
Completing the Square — the verify step in Ex 2, 9.
Degenerate Conics — point circle (Ex 3) and imaginary circle (Ex 4).

Concept Map

Mermaid note for new readers: boxes are outcomes, arrows read "leads to"; the quantity written as g2 plus f2 minus c is exactly $g^{2} + f^{2} - c$ (the circle's radius-squared), and c here always means the circle's constant term.