Visual walkthrough — General form of circle — converting, finding centre and radius

This page rebuilds the parent result — the general form conversion — entirely in pictures. We start from a single dot on graph paper and end holding the centre and radius. Every symbol is earned before it appears.

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Step 1 — What a circle actually is (a dot and a leash)

WHAT. Pick one point on the plane. Call it the centre. Now pin a leash of fixed length to it and walk the end of the leash all the way around. The path the end traces is the circle. The fixed leash-length is the radius.

WHY. Before any algebra, we need the geometric fact every formula must respect: a circle = one centre + one distance. If our final numbers don't produce a single centre and a single distance, we did something wrong. This is our truth-check.

PICTURE. In the figure, the black dot is the centre. The orange leash has length $r$. The blue curve is every place the leash-end can reach.

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Step 2 — Turning "same distance" into an equation

WHAT. Take any point on the circle and call it $(x,y)$ — a roaming point. The straight line from the centre $(h,k)$ to $(x,y)$ is exactly one leash-length, $r$.

WHY. We need to measure that leash. The tool for the length of a straight line between two points is the distance formula, which is just the Pythagoras theorem wearing coordinates:

$\underset{\text{leash length}}{\underbrace{r}}=\sqrt{\underset{\text{across gap, squared}}{\underbrace{(x-h{)}^2}}+\underset{\text{up gap, squared}}{\underbrace{(y-k{)}^2}}}$

The horizontal gap between the roaming point and the centre is $x-h$; the vertical gap is $y-k$. Pythagoras says the diagonal (the leash) squared equals the two gaps squared added together. Squaring both sides to kill the ugly root:

$\underset{{\text{across}}^2}{\underbrace{(x-h{)}^2}}+\underset{{\text{up}}^2}{\underbrace{(y-k{)}^2}}=\underset{{\text{leash}}^2}{\underbrace{r^2}}$

This is the standard form — the circle wearing its centre and radius on its sleeve.

PICTURE. The dashed grey triangle shows the two gaps as legs and the leash as hypotenuse.

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Step 3 — Expanding the tidy form makes a MESS (on purpose)

WHAT. Multiply out both brackets and gather everything on one side.

WHY. Real problems rarely hand you the tidy form. Intersections, loci, and diameter constructions spit out an expanded jumble. To read that, we must first know how the jumble is built. So let's build it.

Expand each square (a square of a difference: $(x-h{)}^2=x^2-2hx+h^2$):

$\underset{\text{from }(x-h{)}^2}{\underbrace{x^2-2hx+h^2}}+\underset{\text{from }(y-k{)}^2}{\underbrace{y^2-2ky+k^2}}=r^2$

Move $r^2$ across and group by type:

$x^2+y^2\underset{\text{linear }x}{\underbrace{-2hx}}\underset{\text{linear }y}{\underbrace{-2ky}}+\underset{\text{one constant}}{\underbrace{(h^2+k^2-r^2)}}=0$

PICTURE. The figure sorts the seven pieces into three coloured bins: the two squared terms (blue), the two linear terms carrying the centre (orange), and the single constant carrying centre and radius (green).

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Step 4 — Naming the disguise: the general form

WHAT. Rename the messy coefficients with fresh letters so we can talk about any such equation, not just one we expanded ourselves.

WHY. We meet the mess first in real problems. Give it standard names $2g$, $2f$, $c$:

$\underset{\text{coeff }1\text{ each}}{\underbrace{x^2+y^2}}+\underset{=-2h}{\underbrace{2g}}x+\underset{=-2k}{\underbrace{2f}}y+\underset{=h^2+k^2-r^2}{\underbrace{c}}=0$

Matching bin-by-bin against Step 3:

• $2g=-2h\Rightarrow g=-h\Rightarrow h=-g$
• $2f=-2k\Rightarrow f=-k\Rightarrow k=-f$
• $c=h^2+k^2-r^2$

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Step 5 — Un-scrambling by completing the square (the reverse move)

WHAT. We undo Step 3. Group the $x$-stuff and $y$-stuff, then rebuild each into a perfect square.

WHY. Completing the square is the only move that forces a lone linear term back into a squared bracket — which is exactly the tidy form we can read. Group first:

$(x^2+2gx)+(y^2+2fy)+c=0$

To turn $x^2+2gx$ into a square, ask: what square starts like this? Answer: $(x+g{)}^2=x^2+2gx+g^2$. It has an extra $g^2$ we never had, so we add it and subtract it (net change zero):

$x^2+2gx=\underset{\text{perfect square}}{\underbrace{(x+g{)}^2}}-\underset{\text{the borrowed bit}}{\underbrace{g^2}}$

Same for $y$: $y^2+2fy=(y+f{)}^2-f^2$.

PICTURE. The figure literally completes a square: the $x^2$ tile plus two $gx$ strips leave a missing corner of area $g^2$; we tile it in (add) and remember we owe it back (subtract).

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Step 6 — Collect, and out pops centre and radius

WHAT. Substitute both completed squares back and push the leftover constants to the right.

WHY. Now the equation looks exactly like Step 2's tidy form, so we just read off the answer.

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

Move the loose constants over:

$\underset{=(x-(-g){)}^2}{\underbrace{(x+g{)}^2}}+\underset{=(y-(-f){)}^2}{\underbrace{(y+f{)}^2}}=\underset{=r^2}{\underbrace{g^2+f^2-c}}$

Compare term-by-term with $(x-h{)}^2+(y-k{)}^2=r^2$:

$\boxed{\text{Centre }=(-g,-f)r=\sqrt{g^2+f^2-c}}$

PICTURE. The figure plots the recovered centre $(-g,-f)$ and draws the radius $\sqrt{g^2+f^2-c}$ — the disguise is off.

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Step 7 — The three fates: real, point, imaginary

WHAT. The quantity under the root, $g^2+f^2-c$, decides everything. Three cases:

$g^2+f^2-c$	What $r$ does	Picture
$>0$	genuine positive radius	a real circle you can draw
$=0$	$r=0$	a single dot — the point circle
$<0$	$r^2$ negative, no real root	imaginary — no points at all

WHY. A radius is a length; lengths can't be imaginary. If the leash-length-squared comes out negative, no real point is that distance from the centre. This is not a bug — it's the algebra honestly reporting "there is no such circle." Watch the circle shrink as $c$ grows: it tightens to the centre (point circle), then vanishes.

PICTURE. Three panels for one centre with growing $c$: fat circle → dot → nothing.

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The one-picture summary

This final figure runs the whole journey left-to-right: leash-circle → tidy form → expanded mess → completing the square → recovered centre and radius, with the decision-branch for the three fates.

Recall Feynman retelling — say it like you'd tell a friend

A circle is just a dot with a leash. If I tell you the dot's spot $(h,k)$ and the leash length $r$, I can write the tidy rule "distance from dot equals $r$," which squared is $(x-h{)}^2+(y-k{)}^2=r^2$. But if I multiply that out, the numbers blend together into a mush: $x^2+y^2+2gx+2fy+c=0$. The dot's location got smeared into the $x$ and $y$ coefficients, and the leash hid inside the lone number $c$. To un-mush it, I complete the square — I patch each group $x^2+2gx$ back into a neat bracket $(x+g{)}^2$, paying back the little corner $g^2$ I borrowed. When the dust clears, the tidy form re-appears and I just read off: centre is $(-g,-f)$ (mind the flip!) and radius is $\sqrt{g^2+f^2-c}$. Last thing: check the number under the root. Positive → real circle. Zero → the circle shrank to a single dot. Negative → the leash length is impossible, so there's no circle at all.

Recall Quick self-test

Centre and radius of $x^2+y^2+8x-6y+10=0$? ::: $g=4,f=-3,c=10$; centre $(-4,3)$; $r=\sqrt{16+9-10}=\sqrt{15}$ Why is the centre $(-g,-f)$ and not $(g,f)$? ::: Because expanding $(x-h{)}^2$ gives $-2hx$, so $2g=-2h\Rightarrow g=-h\Rightarrow h=-g$ What does $g^2+f^2-c=0$ mean geometrically? ::: The radius is zero — a point circle, the limit of a shrinking circle

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Connections

• Standard form of circle — the tidy target we un-scramble to
• Completing the square — the reverse move powering Steps 5–6
• Equation of circle from endpoints of diameter — a common source of the general form
• Tangent to a circle — easiest once centre and radius are recovered
• Family of circles — general form is the natural parameterised shape
• Conic sections general equation — circle is the special case $A=C,B=0$