General form of circle — converting, finding centre and radius

Overview

The general form of a circle is $x^2 + y^2 +2gx + 2fy + c = 0$. This looks messy compared to the standard form $(x-h)^2 + (y-k)^2 = r^2$, but it's how circles often appear after algebraic manipulation. The power lies in conversion: extracting the centre and radius hidden inside those linear terms.

---

[!intuition] Why Does the General Form Exist?

When you expand $(x-h)^2 + (y-k)^2 = r^2$, you get: $x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2$ $x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$

The squared terms have coefficient 1, the linear terms hide the centre, and the constant term hides both centre and radius. Real-world problems (loci, intersections, tangents) often give you the general form directly. You need to reverse-engineer the centre and radius.

---

[!definition] General Form Components

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

• Coefficients of $x^2$ and $y^2$: Must both be 1 (if not, divide through)
• No $xy$ term: Cross-product coefficient must be 0 (pure circle, not ellipse/hyperbola)
• $2g$: Coefficient of $x$ → encodes $-2h$ where $h$ is $x$-coordinate of centre
• $2f$: Coefficient of $y$ → encodes $-2k$ where $k$ is $y$-coordinate of centre
• $c$: Constant → encodes $h^2 + k^2 - r^2$

---

[!formula] Conversion Formulas (Derived from Scratch)

Goal: Extract centre $(-g, -f)$ and radius $r$ from $x^2 + y^2 + 2gx + 2fy + c = 0$.

Derivation: Completing the Square

Why this technique? Because $(x-h)^2$ forces the linear $x$ term to vanish, isolating $h$.

Step 1: Group $x$ terms and $y$ terms: $(x^2 + 2gx) + (y^2 + 2fy) + c = 0$

Step 2: Complete the square for $x$:

• Take half the coefficient of $x$: $\frac{2g}{2} = g$
• Square it: $g^2$
• Add and subtract: $x^2 + 2gx + g^2 - g^2$

$x^2 + 2gx = (x + g)^2 - g^2$

Why? Because $(x+g)^2 = x^2 + 2gx + g^2$. We create the perfect square by "borrowing" $g^2$.

Step 3: Complete the square for $y$ similarly: $y^2 + 2fy = (y + f)^2 - f^2$

Step 4: Substitute back: $(x + g)^2 - g^2 + (y + f)^2 - f^2 + c = 0$ $(x + g)^2 + (y + f)^2 = g^2 + f^2 - c$

Step 5: Compare with $(x - h)^2 + (y - k)^2 = r^2$:

• $(x + g)^2 = (x - (-g))^2 \implies h = -g$
• $(y + f)^2 = (y - (-f))^2 \implies k = -f$
• $r^2 = g^2 + f^2 - c$

Final Result

Why the negative signs? The general form writes $+2gx$, but the standard form has $-2hx$. So $2g = -2h \implies g = -h$.

---

[!example] Example 1: Standard Conversion

Problem: Find centre and radius of $x^2 + y^2 + 6x - 4y - 12 = 0$.

Solution:

Step 1: Identify coefficients by matching $x^2 + y^2 + 2gx + 2fy + c = 0$:

• $2g = 6 \implies g = 3$
• $2f = -4 \implies f = -2$
• $c = -12$

Why this step? Direct pattern matching isolates the parameters.

Step 2: Apply formulas:

• Centre: $C = (-g, -f) = (-3, -(-2)) = (-3, 2)$
• Radius: $r = \sqrt{g^2 + f^2 - c} = \sqrt{9 + 4 - (-12)} = \sqrt{25} = 5$

Why check $g^2 + f^2 - c$? If negative, the "circle" is imaginary (no real points satisfy it).

Answer: Centre $(-3, 2)$, radius $5$.

---

[!example] Example 2: Converting Standard to General

Problem: Convert $(x - 2)^2 + (y + 5)^2 = 16$ to general form.

Solution:

Step 1: Expand the squares: $(x - 2)^2 = x^2 - 4x + 4$ $(y + 5)^2 = y^2 + 10y + 25$

Why expand? The general form has no squared binomials, only $x^2, y^2, x, y, c$ terms.

Step 2: Substitute and collect: $x^2 - 4x + 4 + y^2 + 10y + 25 = 16$ $x^2 + y^2 - 4x + 10y + 29 = 16$ $x^2 + y^2 - 4x + 10y + 13 = 0$

Why move 16? General form sets right side to zero.

Verify: $2g = -4 \implies g = -2$, $2f = 10 \implies f = 5$, $c = 13$.

• Centre: $(-g, -f) = (2, -5)$ ✓
• Radius: $\sqrt{4 + 25 - 13} = \sqrt{16} = 4$ ✓

---

[!example] Example 3: No Real Circle (Imaginary Case)

Problem: Does $x^2 + y^2 + 2x + 4y + 20 = 0$ represent a real circle?

Solution:

Step 1: Extract parameters:

• $2g = 2 \implies g = 1$
• $2f = 4 \implies f = 2$
• $c = 20$

Step 2: Check $r^2 = g^2 + f^2 - c$: $r^2 = 1 + 4 - 20 = -15$

Why is this a problem? Radius squared is negative → no real radius exists.

Answer: This is an imaginary circle (radius is purely imaginary). No real points satisfy it.

Physical intuition: The constant $c$ is too large; the centre and radius data are "contradictory" in real space.

---

[!example] Example 4: Point Circle (Degenerate Case)

Problem: Analyze $x^2 + y^2 - 6x + 4y + 13 = 0$.

Solution:

• $g = -3$, $f = 2$, $c = 13$
• $r^2 = 9 + 4 - 13 = 0 \implies r = 0$

Answer: Centre $(-(-3), -2) = (3, -2)$, radius $0$. This is a point circle — the circle has "collapsed" to a single point. It's the limiting case as a circle shrinks.

---

[!mistake] Common Mistakes & Steel-Manning

Mistake 1: Forgetting the Negative Signs

Wrong approach: $x^2 + y^2 + 6x - 8y + 5 = 0 \implies$ centre is $(6, -8)$.

Why it feels right: You see $+6x$ and think "the centre's $x$-coordinate is 6."

The fix: The formula is $(-g, -f)$, not $(g, f)$. The $+6x$ in general form corresponds to $-2hx$ in standard form, so $h = -3$.

Correct: $2g = 6 \implies g = 3 \implies h = -g = -3$. Centre is $(-3, 4)$.

Mistake 2: Using $c$ Directly as $r^2$

Wrong approach: $x^2 + y^2 + 4x - 2y + 1 = 0 \implies r^2 = 1$.

Why it feels right: The constant term looks like it should be related to radius.

The fix: The constant $c$ also encodes the centre's position: $c = h^2 + k^2 - r^2$. You must compute $r^2 = g^2 + f^2 - c$.

Correct: $g =2, f = -1, c = 1 \implies r^2 = 4 + 1 - 1 = 4 \implies r = 2$.

Mistake 3: Not Checking Validity

Wrong approach: Blindly write $r = \sqrt{g^2 + f^2 - c}$ even when it's negative.

Why it feels right: Algorithms make you rush; you forget to verify.

The fix: Always check $g^2 + f^2 - c \geq 0$. If negative, state "no real circle." If zero, "point circle."

---

[!recall]- Explain It to a 12-Year-Old

Imagine you're hiding a treasure (the centre of a circle) by giving someone clues (the equation). The fancy way to describe a circle is "all points 5 steps away from treasure at location (3, 4)" — that's $(x-3)^2 + (y-4)^2 = 25$.

But if you expand that (like opening a wrapped gift), you get a messy equation like $x^2 + y^2 - 6x - 8y + 0 = 0$. Now the treasure's location (3, 4) and the distance (5 steps) are scrambled into those numbers $-6, -8, 0$.

To find the treasure, you complete the square — it's like re-wrapping the gift. You group the $x$ stuff together: $x^2 - 6x$ becomes $(x-3)^2 - 9$. Same for $y$. Then you see: "Oh! The centre is at (3, 4) and the radius is 5." The $-6$ was secretly $-2\times 3$, hiding the 3. The general form is just the circle in disguise!

---

[!mnemonic] Centre and Radius Quick Recall

"Negative Guys Find Circles"

• Negative: Centre has negative of the coefficients
• Guys: $g$ comes from the $x$ term ($2g$)
• Find: $f$ comes from the $y$ term ($2f$)
• Circles: Use $g^2 + f^2 - c$ to find radius

Formula chant: "Centre is minus-g, minus-f; radius is root of g-squared, f-squared, minus c."

---

Diagram

---

Connections

• Standard form of circle — the target form after completing the square
• Completing the square — the algebraic technique used for conversion
• Equation of circle from endpoints of diameter — often yields general form directly
• Tangent to a circle — easier to find from general form using implicit differentiation
• Family of circles — general form is the natural representation for parameterized circles
• Conic sections general equation — $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$; circle is special case $A=C, B=0$

---

Flashcards

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Circle ka general form kuch ajeb lagta hai pehli baar: $x^2 + y^2 + 2gx + 2fy + c = 0$. Par yeh actually standard form $(x-h)^2 + (y-k)^2 = r^2$ ka hi expanded version hai. Jab ap brackets kholte ho, toh centre aur radius ki information linear terms ($2gx, 2fy$) aur constant ($c$) mein chhip jati hai. General form ka fayda yeh hai ki real-world problems mein circle ka equation aksar isi form mein milta hai — tangent lines, locus problems, ya intersection questions solve karte waqt.

Conversion ka logic simple hai: completing the square technique use karo. $x^2 + 2gx$ ko $(x+g)^2 - g^2$ mein convert karte ho, same $y$ ke liye. Phir equation ban jata hai $(x+g)^2 + (y+f)^2 = g^2 + f^2 - c$. Iska matlab centre hai $(-g, -f)$ — notice the negative signs, bahut important! — aur radius hai $\sqrt{g^2 + f^2 - c}$. Ek trick yad rakho: agar $g^2 + f^2 - c$ negative nikle, toh circle real nahi hai (imaginary). Agar zero ho, toh circle ek point ban gaya (degenerate case).

Exams mein teen chezein puchhenge: general form se centre-radius nikalo, standard form ko general mein convert karo, ya yeh check karo ki circle valid hai ya nahi. Sabse badi mistake yeh hoti hai ki log negative signs bhool jate hain — $2g = 6$ seedha centre $(6, ...)$ likh dete hain, jabki hona chahiye $(-3, ...)$ because centre formula $(-g, -f)$ hai. Dusri mistake: constant $c$ ko directly $r^2$ maan lena. Always formula yad rakho: $r = \sqrt{g^2 + f^2 - c}$. Practice karte raho, yeh conversion bohot useful hai coordinate geometry ke aage ke topics mein!