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

Derivation from First Principles

Start with the definition: A circle with center $(h, k)$ and radius $r$ is the set of all points $(x, y)$ whose distance from the center equals $r$.

Step 1 — Write the distance formula: The distance between any point $(x, y)$ and the center $(h, k)$ is: $d = \sqrt{(x - h)^2 + (y - k)^2}$

WHY? The distance formula comes from the Pythagorean theorem: the horizontal separation is $(x - h)$, vertical is $(y - k)$, and $d^2 = (\text{horizontal})^2 + (\text{vertical})^2$.

Step 2 — Apply the circle condition: For a point to be on the circle, its distance must equal $r$: $\sqrt{(x - h)^2 + (y - k)^2} = r$

Step 3 — Square both sides: $(\sqrt{(x - h)^2 + (y - k)^2})^2 = r^2$ $(x - h)^2 + (y - k)^2 = r^2$

WHY square? To eliminate the square root and get a cleaner algebraic form. Squaring is valid because both sides are positive (distance and radius are always≥ 0).

Special Case: Circle at Origin

When the center is at the origin $(0, 0)$: $(x - 0)^2 + (y - 0)^2 = r^2$ $\boxed{x^2 + y^2 = r^2}$

WHY this simplifies: Subtracting zero does nothing, so $h = k = 0$ disappears from the formula.

Reading Information from Standard Form

1. Center: Look at what's being subtracted from $x$ and $y$

   • $(x - 3)^2$ → $h = 3$
   • $(x + 5)^2 = (x - (-5))^2$ → $h = -5$
   • SIGN FLIP: If you see minus in the equation, the center coordinate is positive

2. Radius: Take the square root of right side

   • If RHS = 25, then $r = \sqrt{25} = 5$
   • MUST be positive: $r > 0$ always

Worked Examples

Solution: Use the standard form directly: $(x - h)^2 + (y - k)^2 = r^2$

Substitute $h = 3$, $k = -2$, $r = 4$: $(x - 3)^2 + (y - (-2))^2 = 4^2$ $(x - 3)^2 + (y + 2)^2 = 16$

Why this step?

• $y - (-2) = y + 2$ (subtracting a negative makes it addition)
• $r^2 = 16$ (we square the radius, not use it directly)

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

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Solution:

Step 1 — Rewrite in standard form: $(x - (-1))^2 + (y - 5)^2 = 49$

Step 2 — Read off $h$, $k$, $r$:

• Center: $(h, k) = (-1, 5)$
• Radius: $r = \sqrt{49} = 7$

Why the sign flip? The equation shows $(x + 1)^2$, which means $(x - (-1))^2$, so $h = -1$, not $+1$. The standard form has minus signs, so a plus in the equation means the center coordinate is negative.

Answer: Center $(-1, 5)$, radius $7$

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Solution:

Substitute $x = 5$, $y = 1$ into the left side: $(5 - 2)^2 + (1 + 3)^2 = 3^2 + 4^2 = 9 + 16 = 25$

Does it equal the right side? Yes, $25 = 25$ ✓

Why this works? If the point is on the circle, its distance from the center must equal $r$. When we substitute and get equality, the distance condition is satisfied.

Answer: Yes, $(5, 1)$ lies on the circle.

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Solution:

Step 1 — Find the radius: The radius is the distance from center $(1, -1)$ to the point $(4, 3)$: $r = \sqrt{(4-1)^2 + (3-(-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Why this step? Since the point $(4, 3)$ is on the circle, its distance from the center is the radius.

Step 2 — Write the equation: $(x - 1)^2 + (y - (-1))^2 = 5^2$ $(x - 1)^2 + (y + 1)^2 = 25$

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

Common Mistakes

Why it feels right: You see the numbers3 and 2, so you just list them.

The fix: The standard form has minus signs: $(x - h)$ and $(y - k)$.

• $(x - 3)$ → $h = +3$ ✓
• $(y + 2) = (y - (-2))$ → $k = -2$ ✓
• Center is $(3, -2)$, not $(3, 2)$

Steel-man: The confusion happens because we mentally "extract" the numbers without tracking the sign structure. Always rewrite with explicit minus: $(y + 2) = (y - (-2))$.

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Why it feels right: The standard form has an $r$ on the right, and you have $r = 5$.

The fix: The standard form is $(x - h)^2 + (y - k)^2 = r^2$, not $r$. You must square the radius.

• If $r = 5$, then $r^2 = 25$
• Equation: $(x - h)^2 + (y - k)^2 = 25$

Steel-man: The formula looks symmetric with squares on both sides of the equation, so it's natural to think "$r$" completes it. But the derivation shows we squared the distance formula: $(\sqrt{\ldots})^2 = r^2$.

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Why it feels right: You're used to "simplifying" expressions, and factored form looks unsimplified.

The fix: Standard form IS the simplified form reading center and radius. Expanding makes it harder to see the circle's properties. Only expand if the problem specifically asks for general form.

Steel-man: In algebra class, "simplified" usually means "no parentheses." But in coordinate geometry, factored form is more informative.

Active Recall Practice

Recall Feynman Explanation (explain to a 12-year-old)

Imagine you're standing at a spot on a field — that's the center of your circle. You have a rope that's exactly $r$ meters long. You tie one end to a stick at the center, hold the other end, and walk around keeping the rope tight. Where you walk traces a circle!

Now, how do we write this as a math equation? Well, if you're at any point $(x, y)$ on your path, the distance from you to the center must equal the rope length $r$. Distance is measured by: "how far right/left" squared, plus "how far up/down" squared, then square root. That's $(x - h)^2 + (y - k)^2$ under a square root.

Set that distance equal to $r$, square both sides to remove the root, and boom: $(x - h)^2 + (y - k)^2 = r^2$. Every point satisfying this equation is on your circle. The equation is just the distance formula saying "stay exactly $r$ away from $(h, k)$!"

$(x - 3)$ → center has $+3$ $(y + 2)$ → center has $-2$

Think: "The equation subtracts the center, so to find the center, flip the sign."

Alternative: "What makes the expression zero?"

• $(x - 3)^2= 0$ when $x = 3$ → center x-coordinate is 3
• $(y + 2)^2 = 0$ when $y = -2$ → center y-coordinate is -2

Connections

• Distance Formula — the foundation of the circle equation
• Pythagorean Theorem — why distance formula works
• General Form of Circle — expanded version: $x^2 + y^2 + Dx + Ey + F = 0$
• Completing the Square — converts general form back to standard form
• Conic Sections — circles are special ellipses with $a = b$
• Equation of Tangent to Circle — uses the circle equation to find tangent lines
• Parametric Equations of Circle — alternative form using $\cos$ and $\sin$

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#flashcards/maths

What is the standard form equation of a circle with center (h, k) and radius r? :: $(x - h)^2 + (y - k)^2 = r^2$

If a circle has center $(-4, 6)$ and radius $3$, what is its equation? :: $(x + 4)^2 + (y - 6)^2 = 9$

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Circle ka equation samajhna bahut simple hai agar tum yeh socho ki circle hai kya — ek center point seab points ko same distance par rakhna. Suppose tumhara center hai $(h, k)$ aur radius hai $r$. Toh agar koi point $(x, y)$ circle par hai, toh uski distance center se exactly $r$ honi chahiye.

Distance formula yad hai? $(x-h)^2 + (y-k)^2$ square root ke andar — yeh tumhe horizontal aur vertical separation ka Pythagorean distance deta hai. Abagar yeh distance $r$ ke barabar ho, matlab $\sqrt{(x-h)^2 + (y-k)^2} = r$, toh point circle par hai. Square root hata do by squaring both sides — mil gaya standard form: $(x-h)^2 + (y-k)^2 = r^2$.

Yeh formula powerful hai kyunki tum seedha dekh sakte ho center kahan hai (sign flip dhyan se: minus ka matlab plus center mein) aur radius kya hai (right side ka square root). Coordinate geometry mein bahut jagah yeh form use hoga — tangents, intersections, distance problems — sab mein. Iska derivation yad rakho because yeh sirf formula nahi hai, yeh circle ki definition ko algebraic language mein translate karna hai.