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

2.3.13 · Maths › Coordinate Geometry

First Principles se Derivation

Definition se shuru karo: Center $(h, k)$ aur radius $r$ wala circle un sabhi points $(x, y)$ ka set hai jinki distance from the center $r$ ke barabar ho.

Step 1 — Distance formula likho: Kisi bhi point $(x, y)$ aur center $(h, k)$ ke beech ki distance hai: $d = (x - h)^{2} + (y - k)^{2}$

KYU? Distance formula Pythagorean theorem se aata hai: horizontal separation $(x - h)$ hai, vertical $(y - k)$ hai, aur $d^{2} = (horizontal)^{2} + (vertical)^{2}$ .

Step 2 — Circle condition apply karo: Koi point circle par hone ke liye, uski distance $r$ ke barabar honi chahiye: $(x - h)^{2} + (y - k)^{2} = r$

Step 3 — Dono sides ko square karo: $((x - h)^{2} + (y - k)^{2})^{2} = r^{2}$ $(x - h)^{2} + (y - k)^{2} = r^{2}$

KYU square karo? Square root hatane ke liye aur ek cleaner algebraic form paane ke liye. Squaring valid hai kyunki dono sides positive hain (distance aur radius hamesha ≥ 0 hote hain).

Special Case: Origin par Circle

Jab center origin $(0, 0)$ par ho: $(x - 0)^{2} + (y - 0)^{2} = r^{2}$ $x^{2} + y^{2} = r^{2}$

KYU yeh simplify hota hai: Zero subtract karne se kuch nahi hota, isliye $h = k = 0$ formula se gayab ho jaata hai.

Standard Form se Information Padhna

Center: Dekho $x$ aur $y$ se kya subtract ho raha hai
- $(x - 3)^{2}$ → $h = 3$
- $(x + 5)^{2} = (x - (- 5))^{2}$ → $h = - 5$
- SIGN FLIP: Agar equation mein minus dikh raha hai, toh center coordinate positive hai
Radius: Right side ka square root lo
- Agar RHS = 25, toh $r = 25 = 5$
- Hamesha positive hona chahiye: $r > 0$ hamesha

Worked Examples

Solution: Standard form directly use karo: $(x - h)^{2} + (y - k)^{2} = r^{2}$

$h = 3$ , $k = - 2$ , $r = 4$ substitute karo: $(x - 3)^{2} + (y - (- 2))^{2} = 4^{2}$ $(x - 3)^{2} + (y + 2)^{2} = 16$

Yeh step kyun?

$y - (- 2) = y + 2$ (negative subtract karne se addition ban jaata hai)
$r^{2} = 16$ (hum radius ko square karte hain, directly use nahi karte)

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

Solution:

Step 1 — Standard form mein rewrite karo: $(x - (- 1))^{2} + (y - 5)^{2} = 49$

Step 2 — $h$ , $k$ , $r$ padhlo:

Center: $(h, k) = (- 1, 5)$
Radius: $r = 49 = 7$

Sign flip kyun? Equation mein $(x + 1)^{2}$ dikh raha hai, matlab $(x - (- 1))^{2}$ , isliye $h = - 1$ hai, $+ 1$ nahi. Standard form mein minus signs hote hain, isliye equation mein plus ka matlab center coordinate negative hai.

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

Solution:

$x = 5$ , $y = 1$ left side mein substitute karo: $(5 - 2)^{2} + (1 + 3)^{2} = 3^{2} + 4^{2} = 9 + 16 = 25$

Kya yeh right side ke barabar hai? Haan, $25 = 25$ ✓

Yeh kyun kaam karta hai? Agar point circle par hai, toh center se uski distance $r$ ke barabar honi chahiye. Jab hum substitute karte hain aur equality milti hai, toh distance condition satisfy ho jaati hai.

Answer: Haan, $(5, 1)$ circle par hai.

Solution:

Step 1 — Radius nikalo: Radius center $(1, - 1)$ se point $(4, 3)$ tak ki distance hai: $r = (4 - 1)^{2} + (3 - (- 1))^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 = 5$

Yeh step kyun? Kyunki point $(4, 3)$ circle par hai, isliye center se uski distance hi radius hai.

Step 2 — Equation likho: $(x - 1)^{2} + (y - (- 1))^{2} = 5^{2}$ $(x - 1)^{2} + (y + 1)^{2} = 25$

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

Common Mistakes

Kyun sahi lagta hai: Tum numbers 3 aur 2 dekhte ho, toh bas unhe likh dete ho.

Fix: Standard form mein minus signs hote hain: $(x - h)$ aur $(y - k)$ .

$(x - 3)$ → $h = + 3$ ✓
$(y + 2) = (y - (- 2))$ → $k = - 2$ ✓
Center $(3, - 2)$ hai, $(3, 2)$ nahi

Steel-man: Confusion isliye hoti hai kyunki hum mentally numbers "extract" karte hain bina sign structure track kiye. Hamesha explicit minus ke saath rewrite karo: $(y + 2) = (y - (- 2))$ .

Kyun sahi lagta hai: Standard form mein right side par $r$ hai, aur tumhare paas $r = 5$ hai.

Fix: Standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ hai, $r$ nahi. Radius ko square karna zaroori hai.

Agar $r = 5$ , toh $r^{2} = 25$
Equation: $(x - h)^{2} + (y - k)^{2} = 25$

Steel-man: Formula equation ke dono sides par squares ke saath symmetric lagta hai, isliye naturally lagta hai ki " $r$ " isse complete karta hai. Lekin derivation dikhata hai ki humne distance formula ko square kiya tha: $(\dots)^{2} = r^{2}$ .

Kyun sahi lagta hai: Tumhari aadat hai expressions ko "simplify" karne ki, aur factored form unsimplified lagta hai.

Fix: Standard form hi simplified form hai — center aur radius padhne ke liye. Expand karne se circle ki properties dekhna mushkil ho jaata hai. Sirf tab expand karo jab problem specifically general form maange.

Steel-man: Algebra class mein "simplified" ka matlab hota hai "no parentheses." Lekin coordinate geometry mein, factored form zyada informative hota hai.

Active Recall Practice

Recall Feynman Explanation (ek 12-saal ke bachche ko samjhao)

Socho tum ek field mein kisi jagah khade ho — yahi tumhare circle ka center hai. Tumhare paas ek rope hai jo exactly $r$ meter lambi hai. Ek end stick se center par baandho, doosra end pakdo, aur rope tight rakhte hue chakkar lagao. Jahan tum chalte ho woh ek circle trace karta hai!

Ab, isse math equation mein kaise likhein? Dekho, agar tum apne path par kisi bhi point $(x, y)$ par ho, toh center se teri distance rope ki length $r$ ke barabar honi chahiye. Distance aise measure hoti hai: "kitna right/left" squared, plus "kitna up/down" squared, phir square root. Yahi $(x - h)^{2} + (y - k)^{2}$ square root ke andar hai.

Woh distance $r$ ke barabar set karo, root hatane ke liye dono sides square karo, aur boom: $(x - h)^{2} + (y - k)^{2} = r^{2}$ . Yeh equation satisfy karne wala har point tumhare circle par hai. Equation bas distance formula hai jo keh raha hai "exactly $r$ door raho $(h, k)$ se!"

$(x - 3)$ → center mein $+ 3$ hai $(y + 2)$ → center mein $- 2$ hai

Socho: "Equation center ko subtract karti hai, isliye center nikaalte waqt sign flip karo."

Alternative: "Kya expression ko zero banata hai?"

$(x - 3)^{2} = 0$ jab $x = 3$ → center ka x-coordinate 3 hai
$(y + 2)^{2} = 0$ jab $y = - 2$ → center ka y-coordinate -2 hai

Connections

Distance Formula — circle equation ki neenv
Pythagorean Theorem — distance formula kyun kaam karta hai
General Form of Circle — expanded version: $x^{2} + y^{2} + D x + E y + F = 0$
Completing the Square — general form ko standard form mein wapas convert karta hai
Conic Sections — circles special ellipses hain jahan $a = b$
Equation of Tangent to Circle — circle equation use karke tangent lines nikalta hai
Parametric Equations of Circle — $cos$ aur $sin$ use karke alternative form

#flashcards/maths

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

For the circle equation

(x - 5)^{2} + (y + 3)^{2} = 16

, what is the center?

(5, - 3)

— note the sign flip:

(y + 3) = (y - (- 3))

means

k = - 3

For the circle equation

(x + 2)^{2} + (y - 7)^{2} = 49

, what is the radius?

r = 49 = 7

What is the equation of a circle centered at the origin with radius

r

x^{2} + y^{2} = r^{2}

How do you find the radius if you know the center

(h, k)

and a point

(x_{1}, y_{1})

on the circle?

Calculate the distance:

r = (x_{1} - h)^{2} + (y_{1} - k)^{2}

Why does the standard form have

r^{2}

instead of

r

on the right side?

Kyunki humne distance equation

(x - h)^{2} + (y - k)^{2} = r

ke dono sides square kiye the square root hatane ke liye

To check if point

(a, b)

is on the circle

(x - h)^{2} + (y - k)^{2} = r^{2}

, what do you do?

x = a

aur

y = b

left side mein substitute karo. Agar yeh

r^{2}

ke barabar aaye, toh point circle par hai.

For

(x - 3)^{2} + (y + 1)^{2} = 25

, what is the center and radius?

Center:

(3, - 1)

, Radius:

5

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

What geometric definition of a circle leads to the standard form equation?

Circle un sabhi points ka set hai jo ek fixed center point

(h, k)

se equal distance (distance

r

) par hain

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

First Principles se Derivation

Special Case: Origin par Circle

Standard Form se Information Padhna

Worked Examples

Common Mistakes

Active Recall Practice

Connections

Concept Map

Deep Dive