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 $r$ (always taken positive). Everything in this topic flows from that one sentence fed through the distance formula, so if you ever forget a result you can rebuild it from scratch.
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 $\sqrt{(x-h)^2+(y-k)^2}=r$ clears the root and gives:
$$(x - h)^2 + (y - k)^2 = r^2$$
When the centre is the origin $(0, 0)$ this collapses to the form you meet most often:
$$x^2 + y^2 = r^2$$
General equation. Expanding the standard form and renaming the constants gives the second-degree equation in which a circle is usually hidden in problems:
$$x^2 + y^2 + 2gx + 2fy + c = 0$$
Matching this term-by-term with the expanded standard form shows that the centre is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$. Two features always betray a circle inside a general second-degree equation: the coefficients of $x^2$ and $y^2$ are equal, and there is no $xy$ term. If the equation comes with a common coefficient (for example $2x^2 + 2y^2 + \dots = 0$), divide through by it first so that the $x^2$ and $y^2$ coefficients are $1$ before reading off $g$, $f$, $c$.
When is it a real circle? The quantity under the root, $g^2 + f^2 - c$, decides whether the equation represents anything real at all. If it is positive you have a genuine circle; if it is zero the "circle" shrinks to the single point $(-g, -f)$ (a point circle); and if it is negative no real point satisfies the equation — an imaginary circle.
Finding the equation from conditions. Most exam questions hand you geometric data and ask for the equation. The strategy depends on what is given:
- Centre and radius given — substitute straight into $(x-h)^2+(y-k)^2=r^2$.
- Centre and a point on the circle — the radius is the distance from the centre to that point.
- Endpoints of a diameter $A$, $B$ — the centre is the midpoint of $AB$ and the radius is half $|AB|$. A neat shortcut is the diameter form $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.
- Three points on the circle — plug each into the general form $x^2+y^2+2gx+2fy+c=0$ to get three linear equations in $g$, $f$, $c$ and solve.
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 — built from one constant distance. When you read 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. That is why completing the square is the master skill for the whole chapter: it converts any messy general equation back into the geometry-revealing standard form, and the same move reappears for the parabola, ellipse and hyperbola when their centres or vertices are shifted off the origin. The condition $g^2 + f^2 - c > 0$ is not an arbitrary rule either; it simply demands that the squared radius be positive, since no real length is the square root of a negative number. Once you read the equation as "squared distance equals squared radius", finding centres and radii stops being a formula hunt and becomes a single, reliable manoeuvre.
Find the equation of the circle with centre $(3, -2)$ and radius $5$.
Solution- 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: $(x-3)^2 + (y+2)^2 = 25$.
Find the centre and radius of the circle $x^2 + y^2 - 6x + 8y - 11 = 0$.
Solution- 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 $6$.
Find the equation of the circle centred at the origin that passes through $(3, 4)$.
Solution- 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: $x^2 + y^2 = 25$.
Find the equation of the circle whose diameter has endpoints $A(1, 2)$ and $B(5, 6)$.
Solution- The centre is the midpoint of the diameter: $\left(\dfrac{1+5}{2}, \dfrac{2+6}{2}\right) = (3, 4)$.
- The radius is 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: $(x-3)^2 + (y-4)^2 = 8$.
Does the equation $x^2 + y^2 + 4x - 6y + 20 = 0$ represent a real circle?
Solution- 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 ($g^2+f^2-c = -7 < 0$).
Find the equation of the circle with centre $(2, -3)$ that passes through the point $(5, 1)$.
Solution- 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: $(x-2)^2 + (y+3)^2 = 25$.
Find the equation of the circle passing through the three points $(0,0)$, $(4,0)$ and $(0,6)$.
Solution- Use the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.
- Through $(0,0)$: $c = 0$.
- Through $(4,0)$: $16 + 8g + c = 0 \Rightarrow 16 + 8g = 0 \Rightarrow g = -2$.
- Through $(0,6)$: $36 + 12f + c = 0 \Rightarrow 36 + 12f = 0 \Rightarrow f = -3$.
- Substitute back: $x^2 + y^2 - 4x - 6y = 0$.
Answer: $x^2 + y^2 - 4x - 6y = 0$, i.e. centre $(2,3)$, radius $\sqrt{13}$.
Find the centre and radius of $2x^2 + 2y^2 - 8x + 12y - 1 = 0$.
Solution- The $x^2$ and $y^2$ coefficients are $2$, so divide the whole equation by $2$ first: $x^2 + y^2 - 4x + 6y - \dfrac{1}{2} = 0$.
- Now $2g = -4$, $2f = 6$, $c = -\dfrac{1}{2}$, giving $g = -2$, $f = 3$.
- Centre $(-g, -f) = (2, -3)$.
- Radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 + \dfrac{1}{2}} = \sqrt{\dfrac{27}{2}} = \dfrac{3\sqrt{6}}{2}$.
Answer: Centre $(2, -3)$, radius $\dfrac{3\sqrt{6}}{2}$.
Find the equation of the circle with $(1,2)$ and $(3,4)$ as the ends of a diameter, using the diameter form.
Solution- The diameter form is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.
- Substitute $(x_1,y_1)=(1,2)$ and $(x_2,y_2)=(3,4)$: $(x-1)(x-3)+(y-2)(y-4)=0$.
- Expand: $x^2 - 4x + 3 + y^2 - 6y + 8 = 0$, i.e. $x^2 + y^2 - 4x - 6y + 11 = 0$.
- Check: centre $(2,3)$, radius $\sqrt{4+9-11}=\sqrt{2}$ — half of $|AB|=\sqrt{8}=2\sqrt2$, as expected.
Answer: $(x-2)^2 + (y-3)^2 = 2$, i.e. $x^2 + y^2 - 4x - 6y + 11 = 0$.
Find the equation of the circle whose centre is $(2, 2)$ and which passes through the point of intersection of the lines $x + y = 4$ and $x - y = 2$.
Solution- First find the point of intersection. Adding the lines: $2x = 6 \Rightarrow x = 3$, then $y = 4 - 3 = 1$, so the point is $(3, 1)$.
- The radius is the distance from $(2,2)$ to $(3,1)$: $r = \sqrt{(3-2)^2 + (1-2)^2} = \sqrt{1+1} = \sqrt{2}$.
- So $r^2 = 2$ and the equation is $(x-2)^2 + (y-2)^2 = 2$.
Answer: $(x-2)^2 + (y-2)^2 = 2$.
Determine whether the point $(4, 1)$ lies inside, on, or outside the circle $x^2 + y^2 - 4x - 2y - 5 = 0$.
Solution- Substitute the point into the left-hand side $S = x^2 + y^2 - 4x - 2y - 5$.
- $S(4,1) = 16 + 1 - 16 - 2 - 5 = -6$.
- $S < 0$ means the point lies inside the circle ($S = 0$ on it, $S > 0$ outside).
Answer: $S(4,1) = -6 < 0$, so $(4, 1)$ lies inside the circle.
A circle has centre on the $x$-axis, passes through the origin, and has radius $4$. Find its equation(s).
Solution- Centre on the $x$-axis means it is $(h, 0)$; the equation is $(x-h)^2 + y^2 = 16$.
- It passes through the origin: $(0-h)^2 + 0 = 16 \Rightarrow h^2 = 16 \Rightarrow h = \pm 4$.
- So there are two such circles, with centres $(4,0)$ and $(-4,0)$.
Answer: $(x-4)^2 + y^2 = 16$ or $(x+4)^2 + y^2 = 16$.