Circles
Equation of a Circle
A circle is the set of all points in a plane that lie at a fixed distance from a fixed point. The fixed point is the centre and the fixed distance is the radius. Everything you need about a circle follows from this single sentence applied through the distance formula.
Standard equation: if the centre is $(h, k)$ and the radius is $r$, a point $(x, y)$ lies on the circle exactly when its distance from the centre equals $r$. Squaring the distance formula gives:
When the centre is the origin $(0, 0)$ this collapses to the form you will use most often:
General form: expanding the standard equation and renaming constants gives the second-degree equation in which a circle is usually hidden:
Comparing this with the expanded standard form, the centre is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$. Notice the two telltale signs of a circle inside a general second-degree equation: the coefficients of $x^2$ and $y^2$ are equal, and there is no $xy$ term.
| Form | Equation | Centre | Radius |
|---|---|---|---|
| Centre $(h,k)$ | $(x-h)^2+(y-k)^2=r^2$ | $(h,k)$ | $r$ |
| Centre at origin | $x^2+y^2=r^2$ | $(0,0)$ | $r$ |
| General form | $x^2+y^2+2gx+2fy+c=0$ | $(-g,-f)$ | $\sqrt{g^2+f^2-c}$ |
The quantity under the root, $g^2 + f^2 - c$, decides whether the equation even represents a real circle: if it is positive you have a genuine circle, if zero the "circle" shrinks to the single point $(-g, -f)$, and if negative no real point satisfies the equation at all.
Deeper Insight — every circle equation is just the distance formula in disguise: The reason a circle has such a tidy equation is that its definition is purely metric — it is built from one constant distance. When you read off the centre as $(-g, -f)$ from the general form, you are really completing the square twice, once in $x$ and once in $y$, to recover the perfect-square brackets that the squared distance always produces. This is why the technique of completing the square is the master skill for the whole chapter: it converts any messy general equation back into the geometry-revealing standard form. The condition $g^2 + f^2 - c > 0$ is not an arbitrary rule either; it simply demands that the squared radius be positive, because no real distance can be the square root of a negative number. Once you see the equation as "squared distance equals squared radius", finding centres and radii stops being a formula hunt and becomes a single, reliable manoeuvre.
- Use the standard form $(x-h)^2 + (y-k)^2 = r^2$ with $h = 3$, $k = -2$, $r = 5$.
- Substitute: $(x-3)^2 + (y-(-2))^2 = 5^2$.
- Simplify: $(x-3)^2 + (y+2)^2 = 25$.
Answer: $(x-3)^2 + (y+2)^2 = 25$.
- Compare with $x^2 + y^2 + 2gx + 2fy + c = 0$: here $2g = -6$, $2f = 8$, $c = -11$.
- So $g = -3$, $f = 4$, giving centre $(-g, -f) = (3, -4)$.
- Radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{9 + 16 - (-11)} = \sqrt{36} = 6$.
Answer: Centre $(3, -4)$, radius $6$.
- Centre is the origin, so the equation is $x^2 + y^2 = r^2$.
- The radius is the distance from $(0,0)$ to $(3,4)$: $r = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
- Hence $x^2 + y^2 = 25$.
Answer: $x^2 + y^2 = 25$.
- The centre is the midpoint of the diameter: $\left(\dfrac{1+5}{2}, \dfrac{2+6}{2}\right) = (3, 4)$.
- The radius is half the diameter, i.e. the distance from the centre to $A$: $r = \sqrt{(3-1)^2 + (4-2)^2} = \sqrt{4+4} = 2\sqrt{2}$.
- So $r^2 = 8$ and the equation is $(x-3)^2 + (y-4)^2 = 8$.
Answer: $(x-3)^2 + (y-4)^2 = 8$.
- Identify $g = 2$, $f = -3$, $c = 20$.
- Compute $g^2 + f^2 - c = 4 + 9 - 20 = -7$.
- Since this is negative, $r^2 < 0$ and no real point satisfies the equation.
Answer: No — it is not a real circle ($g^2+f^2-c = -7 < 0$).
- The radius equals the distance from the centre to the given point.
- $r = \sqrt{(5-2)^2 + (1-(-3))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
- Equation: $(x-2)^2 + (y+3)^2 = 25$.
Answer: $(x-2)^2 + (y+3)^2 = 25$.
- A circle is all points at a fixed distance $r$ (radius) from a fixed point (centre).
- Standard form: $(x-h)^2 + (y-k)^2 = r^2$; centred at origin it is $x^2 + y^2 = r^2$.
- General form $x^2 + y^2 + 2gx + 2fy + c = 0$ has centre $(-g, -f)$ and radius $\sqrt{g^2+f^2-c}$.
- A real circle requires $g^2 + f^2 - c > 0$; equal $x^2$, $y^2$ coefficients and no $xy$ term signal a circle.
- Complete the square to convert any general equation back to standard form.
General Form, Centre and Radius
The general form x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c).
It is a real circle only when g² + f² − c > 0.
- Centre (−g, −f), radius √(g² + f² − c).
- Real circle needs g² + f² − c > 0.
Tangents and Positions
A line is a tangent when its distance from the centre equals the radius, a secant when it is less, and misses when more.
A point lies inside, on, or outside according as x² + y² + 2gx + 2fy + c is negative, zero, or positive.
- Tangent ⟺ distance from centre = radius.
- Sign of the expression places a point in/on/out.