Worked examples — Circle equation — standard form (x−h)² + (y−k)² = r²

This page is a complete catalogue of cases. The parent note built the equation $(x-h{)}^2+(y-k{)}^2=r^2$ from the Distance Formula. Here we hunt down every situation you could meet — good signs, bad signs, zeros, degenerate inputs, a word problem, and an exam trap — and grind each to a finished answer.

Before a single symbol appears in an example, here is the one picture the whole page rests on.

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The scenario matrix

Every problem this topic can throw is one of these cells. The example that kills each cell is named in the last column.

#	Case class	What is tricky about it	Killed by
A	Center in Quadrant I (both $+$), build equation	plain substitution	Ex 1
B	Center with a negative coordinate	the sign flip $(y+2)=(y-(-2))$	Ex 2
C	Center on / straddling an axis ($h=0$ or origin)	terms vanish, don't panic	Ex 3
D	Read back center + radius from a given equation	reverse the sign flip, un-square	Ex 4
E	Point-test: on / inside / outside	comparing LHS to $r^2$, all 3 verdicts	Ex 5
F	Circle through a given point (find $r$ first)	radius = distance, Pythagorean Theorem	Ex 6
G	Degenerate: $r=0$ and "negative $r^2$"	a circle that is a point, or nothing	Ex 7
H	Word problem (real-world leash)	translating words to $(h,k,r)$	Ex 8
I	Exam twist: diameter endpoints given	midpoint = center, half-length = radius	Ex 9

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Example 1 — Cell A: center in Quadrant I

Forecast: guess the equation before reading — where do the $2$, $3$, and $5$ land, and which one gets squared?

1. Write the blank template $(x-h{)}^2+(y-k{)}^2=r^2$. Why this step? Every circle answer is this skeleton with three slots filled — start from the skeleton so no slot is forgotten.
2. Drop in $h=2,k=3$: $(x-2{)}^2+(y-3{)}^2=r^2$. Why this step? $h,k$ are the coordinates you subtract; the center sits in Quadrant I (right and up), so both subtractions are of positive numbers.
3. Square the radius: $r^2=5^2=25$, giving $(x-2{)}^2+(y-3{)}^2=25$. Why this step? The right side is the squared distance, not the distance — the derivation squared both sides to kill the square root.

Verify: the center $(2,3)$ itself should give left side $=0$ (distance zero, you're standing on the pin): $(2-2{)}^2+(3-3{)}^2=0\ne 25$, so $(2,3)$ is inside — correct, the center is always inside. A point one radius up, $(2,8)$: $(0{)}^2+(5{)}^2=25$ ✓ on the circle.

Answer: $(x-2{)}^2+(y-3{)}^2=25$.

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Example 2 — Cell B: a negative center coordinate

Forecast: will the $-2$ show up as $(y-2)$ or $(y+2)$? Commit to an answer first.

1. Template: $(x-h{)}^2+(y-k{)}^2=r^2$.
2. Substitute $h=3,k=-2$: $(x-3{)}^2+(y-(-2){)}^2=r^2$. Why this step? The formula always subtracts $k$. Here $k$ itself is negative, so we subtract a negative.
3. Simplify the double sign: $y-(-2)=y+2$, so $(x-3{)}^2+(y+2{)}^2=r^2$. Why this step? Subtracting a negative is adding — this is the only reason a "$+$" ever appears inside a circle equation.
4. Square the radius: $r^2=16$.

Verify: does the center make each bracket vanish? $x=3\Rightarrow (x-3)=0$ ✓, $y=-2\Rightarrow (y+2)=0$ ✓. Both zero → left side $=0$, confirming $(3,-2)$ is the center.

Answer: $(x-3{)}^2+(y+2{)}^2=16$.

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Example 3 — Cell C: center on an axis / at the origin

Forecast: if $h=0$, does the $x$-bracket vanish or become $x^2$?

(a)

1. Substitute $h=k=0$: $(x-0{)}^2+(y-0{)}^2=3^2$. Why this step? Subtracting zero changes nothing to subtract — the bracket is just $x$.
2. Simplify: $x^2+y^2=9$. Why this step? $(x-0{)}^2=x^2$. This is the origin-centered special case.

(b)

1. $h=0,k=-4$: $(x-0{)}^2+(y-(-4){)}^2=2^2$.
2. Simplify: $x^2+(y+4{)}^2=4$. Why this step? $h=0$ kills the $x$-shift; $k=-4$ becomes a $+4$ inside via the double-sign rule.

Verify (a): point $(3,0)$: $9+0=9$ ✓. Point $(0,-3)$: $0+9=9$ ✓ — the circle reaches distance $3$ in all four axis directions, as a radius-3 circle at the origin must. Verify (b): center $(0,-4)$: $0+0=0$, inside ✓; top point $(0,-2)$: $0+(-2+4{)}^2=4$ ✓ on the circle.

Answers: (a) $x^2+y^2=9$ (b) $x^2+(y+4{)}^2=4$.

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Example 4 — Cell D: read center & radius back out

Forecast: center $x$ — is it $+1$ or $-1$?

1. Rewrite each bracket as a subtraction: $(x+1)=(x-(-1))$ and $(y-5)$ stays. Why this step? Standard form only knows how to subtract; force the equation into that shape so you can just read $h,k$.
2. Read $h=-1,k=5$. Why this step? $h$ is whatever we subtract from $x$; we subtract $-1$, so $h=-1$.
3. Un-square the right side: $r=\sqrt{49}=7$ (positive root only — a leash length can't be negative). Why this step? The right side is $r^2$; radius is its principal square root.

Verify: the center $(-1,5)$ zeroes both brackets: $(-1+1{)}^2+(5-5{)}^2=0$ ✓. A point $7$ to the right, $(6,5)$: $(6+1{)}^2+0=49$ ✓.

Answer: center $(-1,5)$, radius $7$.

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Example 5 — Cell E: is a point on / inside / outside?

This is the yes/no machine run in all three directions.

Forecast: guess each verdict before computing. Which one is exactly on?

For each point compute $L=(x-2{)}^2+(y+3{)}^2$ and compare to $r^2=25$.

1. $P=(5,1)$: $L=(3{)}^2+(4{)}^2=9+16=25$. Since $L=25=r^2$ → on the circle. Why this step? Equal left side and $r^2$ means distance $=r$ exactly.
2. $Q=(2,-1)$: $L=(0{)}^2+(2{)}^2=0+4=4$. Since $4<25$ → inside. Why this step? Smaller than $r^2$ means the point is closer than the leash allows a boundary point to be.
3. $R=(9,-3)$: $L=(7{)}^2+(0{)}^2=49$. Since $49>25$ → outside. Why this step? Larger than $r^2$ means farther than the radius.

Verify: distances directly — $P$: $\sqrt{25}=5=r$ ✓; $Q$: $\sqrt{4}=2<5$ ✓ inside; $R$: $\sqrt{49}=7>5$ ✓ outside. All three cases confirmed.

Answer: $P$ on, $Q$ inside, $R$ outside.

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Example 6 — Cell F: circle through a given point

Forecast: the radius isn't given — where does it come from?

1. Radius = distance from center to the point (that point lies on the leash's end). Why this step? Any point on the circle is exactly one radius from the center — see the red hypotenuse in the figure.
2. Use the Distance Formula / Pythagorean Theorem: $r=\sqrt{(4-1{)}^2+(3-(-1){)}^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5.$ Why this step? Horizontal leg $3$, vertical leg $4$; the leash is the hypotenuse, $r^2=3^2+4^2$.
3. Assemble the equation with $h=1,k=-1,r^2=25$: $(x-1{)}^2+(y+1{)}^2=25.$

Verify: plug the given point $(4,3)$ back: $(4-1{)}^2+(3+1{)}^2=9+16=25$ ✓ — it lands on the circle, as required.

Answer: $(x-1{)}^2+(y+1{)}^2=25$.

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Example 7 — Cell G: degenerate circles

Forecast: can a circle have zero radius? A negative one?

1. (a) Set $r^2=0$, so $r=0$: a leash of length zero pins you to the center — the "circle" collapses to the single point $(4,-2)$. Why this step? $(x-4{)}^2+(y+2{)}^2=0$ is a sum of two squares equal to zero; each square must be $0$, forcing $x=4$ and $y=-2$.
2. (b) $r^2=-9$: the left side is a sum of two squares, which can never be negative. Why this step? Any real square is $\ge 0$, so $(x-4{)}^2+(y+2{)}^2\ge 0>-9$. No point can satisfy it.

Verify (a): the only candidate $(4,-2)$ gives $0+0=0$ ✓, and any nudge, say $(4.1,-2)$, gives $0.01>0$ ✗ — truly a single point. Verify (b): the smallest the left side can be is $0$ (at the center); $0>-9$, so equality is impossible — empty set confirmed.

Answers: (a) the single point $(4,-2)$; (b) no graph (empty set).

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Example 8 — Cell H: a real-world leash (word problem)

Forecast: which number is the center, which is the radius, and is $(0,-2)$ wet or dry?

1. Identify the pieces: the sprinkler is the center $(h,k)=(6,4)$; the throw distance is the radius $r=10$ m. Why this step? "Same distance in every direction from a fixed spot" is the exact definition of a circle — the sprinkler is the pin, the throw is the leash.
2. (i) Boundary equation: $(x-6{)}^2+(y-4{)}^2=10^2=100$. Why this step? Square the radius; units are m${}^2$ on both sides, consistent.
3. (ii) Point-test the flowerbed $(0,-2)$: $L=(0-6{)}^2+(-2-4{)}^2=36+36=72$. Why this step? Compare $L$ to $r^2=100$ — the yes/no machine again.
4. Compare: $72<100$ → the flowerbed is inside, so it gets watered.

Verify: actual distance sprinkler→bed $=\sqrt{72}=\sqrt{72}\approx 8.49$ m $<10$ m ✓ within range. Units check: distances in m, squared distances in m${}^2$, $72{\text{ m}}^2<100{\text{ m}}^2$ ✓.

Answer: (i) $(x-6{)}^2+(y-4{)}^2=100$; (ii) yes, the bed at $(0,-2)$ is inside (about $8.49$ m out, radius $10$ m).

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Example 9 — Cell I: exam twist, diameter endpoints

Forecast: the center isn't handed to you — where is it, and how big is the radius?

1. Center = midpoint of the diameter: $\left(\frac{1+7}{2},\frac{2+10}{2}\right)=(4,6).$ Why this step? A diameter passes through the center, splitting it evenly, so the center is dead centre between the endpoints.
2. Radius = half the diameter length. First the full diameter via Distance Formula: $AB=\sqrt{(7-1{)}^2+(10-2{)}^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.$ So $r=\frac{10}{2}=5$, and $r^2=25$. Why this step? The diameter is twice the radius; halving gives the leash length.
3. Assemble: $(x-4{)}^2+(y-6{)}^2=25$.

Verify: both endpoints must lie on the circle. $A=(1,2)$: $(1-4{)}^2+(2-6{)}^2=9+16=25$ ✓. $B=(7,10)$: $(7-4{)}^2+(10-6{)}^2=9+16=25$ ✓. Both on — center and radius are correct.

Answer: $(x-4{)}^2+(y-6{)}^2=25$.

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Recall

Recall Which cell is each verdict? (point-test)

If plugging a point gives LHS $=r^2$ / $<r^2$ / $>r^2$, the point is… ::: on / inside / outside the circle.

Recall Degenerate radius meanings

What does $r^2=0$ give, and what does $r^2<0$ give? ::: $r^2=0$ → a single point (the center); $r^2<0$ → the empty set (no real points).

Recall Diameter endpoints → equation

Given endpoints of a diameter, how do you get $(h,k)$ and $r$? ::: center = midpoint of the endpoints; $r=\frac{1}{2}\times$ (distance between endpoints).

Connections

• Distance Formula — powers Examples 6 and 9.
• Pythagorean Theorem — the legs-and-hypotenuse picture behind the radius.
• Completing the Square — the "what makes the bracket zero?" trick for the sign flip.
• General Form of Circle — expand any answer here to see it.
• Conic Sections, Equation of Tangent to Circle, Parametric Equations of Circle — where these circles go next.