General form of circle — converting, finding centre and radius: Worked examples

This page is the drill floor. The parent note taught you the machine: from $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , the centre is $(- g, - f)$ and the radius is $g^{2} + f^{2} - c$ . Here we run that machine against every kind of input it could ever meet — so you never hit a surprise in an exam.

Before any example, let us name the two forms we keep translating between, plus the ingredient we re-use.

Figure — General form of circle — converting, finding centre and radius

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. Each row is a distinct kind of situation; the worked examples below are labelled with the cell they hit, and together they touch every cell.

Cell	What makes it different	Danger it tests
A. Sign salad	Mixed $+$ and $-$ signs in the $x$ , $y$ , $c$ terms	Losing a minus sign in $(- g, - f)$
B. All-positive centre	Both linear coefficients negative → centre in Quadrant I	Trusting the sign you see
C. Centre on an axis	A linear term is missing (its coefficient is $0$ )	Realising $g = 0$ or $f = 0$ is allowed
D. Point circle	$D = 0$ exactly	Reporting radius $0$ , not "no circle"
E. Imaginary circle	$D < 0$	Accepting that no real points exist — there is no curve to draw
F. Coefficients not 1	$x^{2}, y^{2}$ have a coefficient like $2$ or $3$	Must divide the whole equation first
G. Standard → General	Reverse direction (expand the squares)	Sign errors when moving the constant across the $=$
H. Word problem	Real-world story hides a circle	Building the equation yourself
I. Exam twist	Unknown constant $k$ , solve for a condition	Turning "point circle" into an equation

Cell A + B — the sign salad and the all-positive centre

Step 1 — Read off the raw coefficients. $2 g = - 10$ , $2 f = + 6$ , $c = - 2$ . Why this step? The formulas are written in terms of $g, f, c$ , so we must first translate the equation into those exact three numbers before anything else can happen.

Step 2 — Halve to get $g$ and $f$ . $g = \frac{- 10}{2} = - 5, f = \frac{6}{2} = 3.$ Why this step? The parameter is $g$ , but the equation shows $2 g$ . Dividing by $2$ undoes that doubling. This is where sign mistakes are born, so go slow.

Step 3 — Flip the signs for the centre. $C = (- g, - f) = (- (- 5), - (3)) = (5, - 3) .$ Why this step? Recall the two forms from the top of the page. Standard form has $(x - h)^{2}$ , which expands to give a $- 2 h x$ term, but general form writes $+ 2 g x$ . Matching those two $x$ -terms forces $2 g = - 2 h$ , i.e. $h = - g$ (and likewise $k = - f$ ). So the centre coordinates $(h, k)$ are the negatives of $g$ and $f$ . That negation is Cell A's whole trap.

Step 4 — Radius via $D$ . $D = g^{2} + f^{2} - c = (- 5)^{2} + 3^{2} - (- 2) = 25 + 9 + 2 = 36 > 0.$ $r = 36 = 6.$ Why this step? $D > 0$ first confirms a real circle exists; only then does taking the square root make sense.

Verify: Plug the point $(5, - 3 + 6) = (5, 3)$ (top of the circle) into the original equation: $25 + 9 - 50 + 18 - 2 = 0.$ ✓ It sits on the curve, exactly $r = 6$ above the centre.

Answer: Centre $(5, - 3)$ , radius $6$ . (Cell B note: centre- $x$ came out positive from a negative $x$ -term — exactly the flip we forecast.)

Cell C — the centre sits on an axis

Step 1 — Spot the invisible coefficient. There is no $x$ term, which means $2 g = 0$ , so $g = 0$ . Then $2 f = - 8 \Rightarrow f = - 4$ , and $c = - 9$ . Why this step? A "missing" term is not missing information — it is the coefficient $0$ shouting at you. Ignoring it would leave $g$ undefined; naming it $0$ keeps the machine running.

Step 2 — Centre. $C = (- g, - f) = (0, 4) .$ Why this step? With $g = 0$ , the negation does nothing to the $x$ -coordinate, so the centre lands exactly on the $y$ -axis. That is the geometric meaning of a missing $x$ -term.

Step 3 — Radius. $D = 0^{2} + (- 4)^{2} - (- 9) = 16 + 9 = 25, r = 5.$

Verify: The lowest point of the circle is $(0, 4 - 5) = (0, - 1)$ . Substitute: $0 + 1 - 8 (- 1) - 9 = 1 + 8 - 9 = 0.$ ✓

Answer: Centre $(0, 4)$ on the $y$ -axis, radius $5$ .

The figure below draws exactly this circle. Notice the violet dot — the centre $(0, 4)$ — sits on the vertical $y$ -axis, which is the visual signature of $g = 0$ . The orange arrow is the radius $r = 5$ dropping straight down to the lowest point $(0, - 1)$ that we tested in the Verify step.

Cell D — the point circle ( $D = 0$ )

Step 1 — Parameters. $2 g = 4 \Rightarrow g = 2$ ; $2 f = - 2 \Rightarrow f = - 1$ ; $c = 5$ .

Step 2 — The critical discriminant. $D = 2^{2} + (- 1)^{2} - 5 = 4 + 1 - 5 = 0.$ Why this step? $D$ is $r^{2}$ . Getting $D = 0$ is not an error — it is a special answer: the radius is $0$ .

Step 3 — Interpret $r = 0$ . The centre is $(- g, - f) = (- 2, 1)$ and $r = 0 = 0$ . A circle of zero radius is just its own centre: the "circle" has collapsed to the single point $(- 2, 1)$ . Why this step? $r = 0$ means "all points $0$ distance from the centre" — and only the centre itself is $0$ away from itself.

Verify: Rewrite by completing the square: $(x + 2)^{2} + (y - 1)^{2} = 0$ . A sum of two squares equals $0$ only when each square is $0$ , forcing $x = - 2$ and $y = 1$ — a lone point. ✓

Answer: A point circle at $(- 2, 1)$ ; radius $0$ .

Cell E — the imaginary circle ( $D < 0$ )

Step 1 — Parameters. $g = - 1$ , $f = 2$ , $c = 9$ .

Step 2 — Discriminant. $D = (- 1)^{2} + 2^{2} - 9 = 1 + 4 - 9 = - 4 < 0.$

Step 3 — Interpret $D < 0$ . $r^{2} = - 4$ , so $r = - 4$ , which is not a real number. No real point $(x, y)$ can satisfy the equation. Why this step? A distance squared can never be negative for real points. When the algebra demands it, geometry refuses: there is simply no curve to draw.

Verify: Completed-square form is $(x - 1)^{2} + (y + 2)^{2} = - 4$ . The left side is a sum of squares $\geq 0$ ; it can never equal $- 4$ . No solutions exist. ✓

Answer: No real circle (an imaginary circle). The large $c$ over-shrinks the radius past zero.

Cell F — coefficients of $x^{2}$ and $y^{2}$ are not 1

Step 1 — Divide the whole equation by $3$ . $x^{2} + y^{2} - 4 x + 6 y - 5 = 0.$ Why this step? Completing the square only produces $(x + g)^{2}$ cleanly when $x^{2}$ stands alone with coefficient $1$ . If the coefficient were $3$ , the derivation that gave us $(- g, - f)$ would be wrong. Dividing every term by $3$ restores the required form without changing the solution set (dividing an equation by a nonzero number keeps the same points).

Step 2 — Now read parameters. $g = - 2$ , $f = 3$ , $c = - 5$ .

Step 3 — Centre and radius. $C = (- g, - f) = (2, - 3) .$ $D = (- 2)^{2} + 3^{2} - (- 5) = 4 + 9 + 5 = 18, r = 18 = 32 .$

Verify: Multiply the completed form back: $(x - 2)^{2} + (y + 3)^{2} = 18 \Rightarrow x^{2} - 4 x + 4 + y^{2} + 6 y + 9 = 18 \Rightarrow x^{2} + y^{2} - 4 x + 6 y - 5 = 0$ . Multiply by $3$ to recover the original. ✓ And $r^{2} = 18$ so $r = 32 \approx 4.243$ .

Answer: Centre $(2, - 3)$ , radius $32$ .

Cell G — standard form back to general form

Step 1 — Expand each bracket. $(x + 4)^{2} = x^{2} + 8 x + 16, (y - 1)^{2} = y^{2} - 2 y + 1.$ Why this step? General form contains only $x^{2}, y^{2}, x, y$ and a constant — no squared brackets. Expanding is the tool from Completing the square run in reverse.

Step 2 — Add and set equal to $20$ . $x^{2} + y^{2} + 8 x - 2 y + 17 = 20.$

Step 3 — Move $20$ across. $x^{2} + y^{2} + 8 x - 2 y - 3 = 0.$ Why this step? General form always has $0$ on the right. To move $+ 20$ from the right to the left you subtract $20$ from both sides, which flips its sign: $17 - 20 = - 3$ . This sign flip is Cell G's classic slip — students who write $+ 37$ or leave $+ 17$ have forgotten that crossing the $=$ changes the sign.

Verify (round-trip): From this, $g = 4$ , $f = - 1$ , $c = - 3$ . Centre $(- 4, 1)$ ✓ matches the standard form. $D = 16 + 1 + 3 = 20 = r^{2}$ ✓ matches the right-hand side. The two forms agree perfectly.

Answer: $x^{2} + y^{2} + 8 x - 2 y - 3 = 0$ , with $g = 4, f = - 1, c = - 3$ .

Cell H — a real-world word problem

Step 1 — Build the standard form from the story. Centre $(3, 2)$ , radius $5$ : $(x - 3)^{2} + (y - 2)^{2} = 25.$ Why this step? "Within $5$ m of $(3, 2)$ " is literally the definition of a circle with that centre and radius — the standard form is the direct translation of the sentence.

Step 2 — Expand to general form. $x^{2} - 6 x + 9 + y^{2} - 4 y + 4 = 25$ $x^{2} + y^{2} - 6 x - 4 y - 12 = 0.$ Why this step? The software wants general form, so we run Cell G's expansion.

Step 3 — Test the corner post $(7, 5)$ . $7^{2} + 5^{2} - 6 (7) - 4 (5) - 12 = 49 + 25 - 42 - 20 - 12 = 0.$ Why this step? A point lies on the circle exactly when its coordinates make the equation equal $0$ . Zero means "on the boundary".

Verify (distance check): Distance from $(3, 2)$ to $(7, 5)$ is $(7 - 3)^{2} + (5 - 2)^{2} = 16 + 9 = 25 = 5$ m $= r$ . ✓ The post is exactly on the watered edge.

Answer: Stored equation $x^{2} + y^{2} - 6 x - 4 y - 12 = 0$ ; the post at $(7, 5)$ lies precisely on the boundary.

The figure shows the sprinkler picture. The violet marker is the nozzle (the centre $(3, 2)$ ), the shaded orange disc is the watered region, and the navy square is the corner post $(7, 5)$ . The orange arrow is the $5$ m radius reaching from nozzle to post — because that arrow lands exactly on the boundary, the post is neither inside nor outside but right on the edge, matching our Verify calculation.

Cell I — the exam twist (solve for an unknown constant)

Step 1 — Parameters in terms of $k$ . $g = 3$ , $f = - 4$ , $c = k$ .

Step 2 — Write the point-circle condition. Point circle $\Leftrightarrow D = 0$ : $g^{2} + f^{2} - c = 0 \Rightarrow 9 + 16 - k = 0.$ Why this step? "Point circle" is not a shape to draw but a condition to impose. Setting $r^{2} = 0$ turns the word into a solvable equation.

Step 3 — Solve for $k$ . $25 - k = 0 \Rightarrow k = 25.$

Step 4 — Explore the neighbours. $D = 25 - k$ .

$k < 25$ : $D > 0$ → a real circle of radius $25 - k$ .
$k = 25$ : $D = 0$ → point circle at $(- 3, 4)$ .
$k > 25$ : $D < 0$ → no real circle (imaginary).

Why this step? Cell I always wants the full behaviour, not one number. As $k$ climbs, the radius $25 - k$ shrinks to $0$ and then goes imaginary — a clean limiting process.

Verify: At $k = 25$ : $(x + 3)^{2} + (y - 4)^{2} = 0$ , forcing $(x, y) = (- 3, 4)$ , a single point. ✓ At $k = 24$ : $D = 1$ , $r = 1$ , a tiny real circle. At $k = 26$ : $D = - 1$ , no real points. ✓

Answer: $k = 25$ gives a point circle at $(- 3, 4)$ ; smaller $k$ gives a real circle, larger $k$ gives no real circle.

[!recall]- One-line recall for each cell

Which sign do you apply to

g

and

f

to get the centre?

Negative — the centre is

(- g, - f)

A term is missing (no

x

term). What is

g

g = 0

; the centre lies on the

y

-axis.

D = 0

means the graph is a…?

Point circle (radius

0

) — a single point at the centre.

D < 0

means…?

No real circle; there are no real points on it, so refuse to take the square root and say so.

Coefficients of

x^{2}, y^{2}

are

3

. First move?

Divide the entire equation by

3

"Point circle" as an equation in an unknown

k

becomes…?

Set

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

and solve.

Connections

General form of circle — converting, finding centre and radius — the parent machine drilled here
Standard form of circle — the target of every conversion (Cells D, F, G)
Completing the square — the engine behind converting both directions
Equation of circle from endpoints of diameter — another route that lands you in general form
Tangent to a circle — the "on the boundary $\Rightarrow$ equation $= 0$ " idea (Cell H) extends here
Family of circles — the unknown- $k$ twist of Cell I generalises to a whole family
Conic sections general equation — Cell F's "divide first" is the circle's special place among conics