MISSION · Geometry
The Equation of a Circle
Pythagoras, applied to every point at once
LEVEL 1 · EXPLORE
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LEVEL 2 · LEARN
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The idea in one sentence
A circle is every point a fixed distance from a fixed centre, and writing that sentence with the distance formula gives its equation.
\[(x - h)^2 + (y - k)^2 = r^2\]Where it comes from
Take a centre \((h, k)\) and a point \((x, y)\) on the circle. The distance between them is the radius, so by the distance formula:
\[\sqrt{(x-h)^2 + (y-k)^2} = r\]Square both sides and you have the standard form. That’s the whole derivation — the equation is the Pythagorean theorem asserted about every point of the circle simultaneously.
Centred at the origin it simplifies to \(x^2 + y^2 = r^2\), which is the version worth recognising instantly.
Reading it off
The signs are the one thing to be careful about. In \((x - h)\), the centre coordinate is \(+h\) — the minus sign is part of the form, not the number.
\[(x - 3)^2 + (y + 2)^2 = 25\]Centre \((3, -2)\), radius 5. The \(+2\) means \(y - (-2)\), so the \(y\) coordinate is negative. And the right-hand side is \(r^2\), so the radius is its square root, not 25.
The expanded form
Multiply the brackets out and you get something less readable:
\[x^2 + y^2 - 6x + 4y - 12 = 0\]Same circle, hidden. To recover the centre and radius, complete the square on the \(x\) terms and the \(y\) terms separately:
\((x^2 - 6x) + (y^2 + 4y) = 12\) \((x - 3)^2 - 9 + (y + 2)^2 - 4 = 12\) \((x - 3)^2 + (y + 2)^2 = 25\)
Recognising a circle in expanded form is the giveaway that it’s a circle at all: equal coefficients on \(x^2\) and \(y^2\), both positive, and no \(xy\) term.
What it lets you do
Once the equation is written, geometric questions become algebra:
- Is a point inside, on, or outside? Put it in. Less than \(r^2\) means inside.
- Where does a line cross the circle? Substitute the line’s equation and solve the resulting quadratic. Two solutions means it cuts through, one means it’s a tangent, none means it misses — and the discriminant tells you which without solving.
- Find the tangent at a point. The tangent is perpendicular to the radius there, so its gradient is the negative reciprocal of the radius’s gradient.
Common mix-ups
- Reading the centre’s signs backwards. \((x + 4)\) means the centre is at \(-4\).
- Using \(r^2\) as the radius. Take the square root.
- Forgetting to balance when completing the square — what you add inside must come back out.
- Assuming any second-degree equation is a circle. The \(x^2\) and \(y^2\) coefficients must match.
Check yourself
Give the centre and radius of \((x + 1)^2 + (y - 4)^2 = 9\), and say whether \((2, 4)\) is on it.
Centre \((-1, 4)\), radius 3. Substituting: \((2+1)^2 + (4-4)^2 = 9\), which equals 9 — so the point is exactly on the circle.
LEVEL 3 · REFLECT
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