Where Does the Equation Come From?#
A circle is the set of all points that are the same distance (the radius $r$) from a fixed point (the centre). That’s it — that’s the entire definition. The equation of a circle is just the distance formula applied to this definition.
Deriving the Equation#
Take any point $P(x; y)$ on a circle with centre $C(a; b)$ and radius $r$.
The distance from $P$ to $C$ is:
$$\text{distance} = \sqrt{(x - a)^2 + (y - b)^2}$$Since every point on the circle is exactly $r$ units from the centre:
$$\sqrt{(x - a)^2 + (y - b)^2} = r$$Square both sides:
$$\boxed{(x - a)^2 + (y - b)^2 = r^2}$$This is the standard form of the equation of a circle.
Key insight: The equation is literally saying “the distance from $(x; y)$ to $(a; b)$ equals $r$.” Every point $(x; y)$ that satisfies this equation lies on the circle. Nothing more, nothing less.
1. Circle Centred at the Origin#
When the centre is at $(0; 0)$, the formula simplifies:
$$(x - 0)^2 + (y - 0)^2 = r^2$$$$\boxed{x^2 + y^2 = r^2}$$Worked Example 1 — Circle at the Origin#
$$x^2 + y^2 = 25$$Write the equation of the circle with centre $(0; 0)$ and radius $5$.
Check: Does $(3; 4)$ lie on this circle? $3^2 + 4^2 = 9 + 16 = 25 = r^2\;\checkmark$
Does $(1; 2)$? $1 + 4 = 5 \neq 25$ — no, it’s inside the circle.
2. Circle with a Shifted Centre#
When the centre is at $(a; b)$:
$$\boxed{(x - a)^2 + (y - b)^2 = r^2}$$The Sign Convention#
The signs in the brackets are always subtraction. This means:
| Centre | Equation | What happens in the brackets |
|---|---|---|
| $(3; 2)$ | $(x - 3)^2 + (y - 2)^2 = r^2$ | Positive coords → minus signs |
| $(-4; 1)$ | $(x + 4)^2 + (y - 1)^2 = r^2$ | $x - (-4) = x + 4$ |
| $(2; -5)$ | $(x - 2)^2 + (y + 5)^2 = r^2$ | $y - (-5) = y + 5$ |
| $(-1; -3)$ | $(x + 1)^2 + (y + 3)^2 = r^2$ | Both negative → both plus signs |
Reading the centre from the equation: The centre coordinates are always the opposite sign of what you see in the brackets. If you see $(x + 4)^2$, the $x$-coordinate of the centre is $-4$.
Worked Example 2 — Writing the Equation#
$$(x - (-2))^2 + (y - 5)^2 = 7^2$$$$(x + 2)^2 + (y - 5)^2 = 49$$Write the equation of the circle with centre $(-2; 5)$ and radius $7$.
Worked Example 3 — Reading Centre and Radius#
State the centre and radius of $(x - 3)^2 + (y + 1)^2 = 16$.
Centre: $(3; -1)$ (flip the signs: $-3 \to 3$, $+1 \to -1$)
Radius: $r = \sqrt{16} = 4$
3. Completing the Square — Converting to Standard Form#
Exam questions often give the equation in expanded (general) form:
$$x^2 + y^2 + Dx + Ey + F = 0$$You must convert this to standard form to read off the centre and radius.
The Method#
- Group the $x$-terms together and $y$-terms together, move the constant to the RHS
- Complete the square for $x$: take half the coefficient of $x$, square it, add to both sides
- Complete the square for $y$: same process
- Write in standard form and read off centre and radius
Worked Example 4 — Full Completing the Square#
Find the centre and radius of $x^2 + y^2 + 4x - 6y + 4 = 0$.
Step 1 — Group and move constant:
$$(x^2 + 4x) + (y^2 - 6y) = -4$$Step 2 — Complete the square for $x$:
Half of $4$ is $2$. Square it: $2^2 = 4$. Add $4$ to both sides.
$$(x^2 + 4x + 4) + (y^2 - 6y) = -4 + 4$$Step 3 — Complete the square for $y$:
Half of $-6$ is $-3$. Square it: $(-3)^2 = 9$. Add $9$ to both sides.
$$(x^2 + 4x + 4) + (y^2 - 6y + 9) = -4 + 4 + 9$$Step 4 — Write in standard form:
$$(x + 2)^2 + (y - 3)^2 = 9$$Centre: $(-2; 3)$ and Radius: $r = \sqrt{9} = 3$
Worked Example 5 — With Negative Constant#
$$(x^2 - 10x) + (y^2 + 2y) = -17$$Find the centre and radius of $x^2 + y^2 - 10x + 2y + 17 = 0$.
Complete the square for $x$: $\left(\frac{-10}{2}\right)^2 = 25$
Complete the square for $y$: $\left(\frac{2}{2}\right)^2 = 1$
$$(x^2 - 10x + 25) + (y^2 + 2y + 1) = -17 + 25 + 1$$$$(x - 5)^2 + (y + 1)^2 = 9$$Centre: $(5; -1)$ and Radius: $r = 3$
When It’s NOT a Circle#
If after completing the square, the RHS is zero or negative, the equation does not represent a circle:
- $r^2 = 0$: a single point (a “circle” with zero radius)
- $r^2 < 0$: no real graph exists (impossible — distance can’t be negative)
4. Determining the Position of a Point#
Given a circle $(x - a)^2 + (y - b)^2 = r^2$ and a point $P(x_1; y_1)$, substitute the point into the LHS:
$$d^2 = (x_1 - a)^2 + (y_1 - b)^2$$| Compare $d^2$ to $r^2$ | Position of $P$ | Meaning |
|---|---|---|
| $d^2 = r^2$ | On the circle | Distance from centre = radius |
| $d^2 < r^2$ | Inside the circle | Distance from centre < radius |
| $d^2 > r^2$ | Outside the circle | Distance from centre > radius |
Worked Example 6 — Point Position#
Determine whether $A(1; 4)$, $B(3; 7)$, and $C(6; 3)$ are inside, on, or outside the circle $(x - 2)^2 + (y - 5)^2 = 10$.
Centre: $(2; 5)$, $r^2 = 10$
Point $A(1; 4)$: $(1-2)^2 + (4-5)^2 = 1 + 1 = 2 < 10$ → Inside
Point $B(3; 7)$: $(3-2)^2 + (7-5)^2 = 1 + 4 = 5 < 10$ → Inside
Point $C(6; 3)$: $(6-2)^2 + (3-5)^2 = 16 + 4 = 20 > 10$ → Outside
5. Finding the Equation — Common Exam Scenarios#
Scenario 1: Given Centre and a Point on the Circle#
Find the equation of the circle with centre $(1; -2)$ passing through $(4; 2)$.
Calculate $r^2$ using the distance formula:
$$r^2 = (4 - 1)^2 + (2 - (-2))^2 = 9 + 16 = 25$$$$\boxed{(x - 1)^2 + (y + 2)^2 = 25}$$Scenario 2: Given the Endpoints of a Diameter#
$A(-1; 3)$ and $B(5; -1)$ are endpoints of a diameter. Find the equation of the circle.
Step 1 — Centre is the midpoint of $AB$:
$$C = \left(\frac{-1 + 5}{2};\;\frac{3 + (-1)}{2}\right) = (2; 1)$$Step 2 — Radius is half the diameter (or distance from centre to either endpoint):
$$r^2 = (5 - 2)^2 + (-1 - 1)^2 = 9 + 4 = 13$$$$\boxed{(x - 2)^2 + (y - 1)^2 = 13}$$Scenario 3: Given Centre and Tangent Line#
The circle has centre $(3; -1)$ and the line $y = 2x + 1$ is a tangent. Find the equation.
The radius equals the perpendicular distance from the centre to the tangent line.
Rewrite the line: $2x - y + 1 = 0$
$$r = \frac{|2(3) - (-1) + 1|}{\sqrt{2^2 + (-1)^2}} = \frac{|6 + 1 + 1|}{\sqrt{5}} = \frac{8}{\sqrt{5}}$$$$r^2 = \frac{64}{5}$$$$\boxed{(x - 3)^2 + (y + 1)^2 = \frac{64}{5}}$$6. Intersection of a Line and a Circle#
To find where a line meets a circle, substitute the line equation into the circle equation.
Worked Example 7 — Line Meets Circle#
Find the points of intersection of $y = x + 1$ and $x^2 + y^2 = 25$.
Substitute $y = x + 1$:
$$x^2 + (x + 1)^2 = 25$$$$x^2 + x^2 + 2x + 1 = 25$$$$2x^2 + 2x - 24 = 0$$$$x^2 + x - 12 = 0$$$$(x + 4)(x - 3) = 0$$$x = -4 \Rightarrow y = -3$ and $x = 3 \Rightarrow y = 4$
Intersection points: $(-4; -3)$ and $(3; 4)$
What the discriminant tells you:
- $\Delta > 0$: Line cuts the circle at 2 points (secant)
- $\Delta = 0$: Line touches the circle at 1 point (tangent)
- $\Delta < 0$: Line misses the circle entirely
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Reading centre as $(2; 3)$ from $(x + 2)^2 + (y - 3)^2$ | $(x + 2)$ means $x - (-2)$, so $a = -2$ | Always flip the sign you see |
| Forgetting $r^2$, writing $r$ | $(x-1)^2 + (y-2)^2 = 5$ has $r = \sqrt{5}$, not $r = 5$ | The equation has $r^2$ on the RHS |
| Completing the square — not adding to both sides | Changes the equation | Whatever you add to the LHS, add to the RHS |
| Assuming $r^2$ is always a perfect square | $r^2 = 13$ is perfectly valid | Leave it as $13$, don’t try to simplify |
| Not checking if $r^2 > 0$ | If completing the square gives $r^2 \leq 0$, it’s not a circle | Always verify the RHS is positive |
💡 Pro Tips for Exams#
1. The Distance Formula is Everything#
The circle equation is the distance formula. If you ever forget it, just write “distance from $(x; y)$ to centre = $r$” and square both sides. You’ll derive the formula on the spot.
2. Quick Centre-Reading#
When you see $(x + 3)^2 + (y - 7)^2 = 20$, train yourself to instantly say: “Centre $(-3; 7)$, radius $\sqrt{20} = 2\sqrt{5}$.” This should be as automatic as reading a word.
3. Completing the Square Checklist#
Before you start, make sure:
- The coefficients of $x^2$ and $y^2$ are both 1 (if not, divide the whole equation first)
- There is no $xy$ term (if there is, it’s not a circle)