Conic Sections • Topic 1 of 3

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:

$$(x - h)^2 + (y - k)^2 = r^2$$

When the centre is the origin $(0, 0)$ this collapses to the form you will use most often:

$$x^2 + y^2 = r^2$$

General form: expanding the standard equation and renaming constants gives the second-degree equation in which a circle is usually hidden:

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

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.

FormEquationCentreRadius
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.

A circle with centre and radius labelled Circle: centre (h, k), radius r x y (h, k) r (x, y) From general form to centre and radius Reading the general form x² + y² + 2gx + 2fy + c = 0general form centre (−g, −f)r = √(g² + f² − c)
1
Worked Example
Find the equation of the circle with centre $(3, -2)$ and radius $5$.
Solution
  1. Use the standard form $(x-h)^2 + (y-k)^2 = r^2$ with $h = 3$, $k = -2$, $r = 5$.
  2. Substitute: $(x-3)^2 + (y-(-2))^2 = 5^2$.
  3. Simplify: $(x-3)^2 + (y+2)^2 = 25$.

Answer: $(x-3)^2 + (y+2)^2 = 25$.

2
Worked Example
Find the centre and radius of the circle $x^2 + y^2 - 6x + 8y - 11 = 0$.
Solution
  1. Compare with $x^2 + y^2 + 2gx + 2fy + c = 0$: here $2g = -6$, $2f = 8$, $c = -11$.
  2. So $g = -3$, $f = 4$, giving centre $(-g, -f) = (3, -4)$.
  3. Radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{9 + 16 - (-11)} = \sqrt{36} = 6$.

Answer: Centre $(3, -4)$, radius $6$.

3
Worked Example
Find the equation of the circle centred at the origin that passes through $(3, 4)$.
Solution
  1. Centre is the origin, so the equation is $x^2 + y^2 = r^2$.
  2. The radius is the distance from $(0,0)$ to $(3,4)$: $r = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
  3. Hence $x^2 + y^2 = 25$.

Answer: $x^2 + y^2 = 25$.

4
Worked Example
Find the equation of the circle whose diameter has endpoints $A(1, 2)$ and $B(5, 6)$.
Solution
  1. The centre is the midpoint of the diameter: $\left(\dfrac{1+5}{2}, \dfrac{2+6}{2}\right) = (3, 4)$.
  2. The radius is half the diameter, i.e. the distance from the centre to $A$: $r = \sqrt{(3-1)^2 + (4-2)^2} = \sqrt{4+4} = 2\sqrt{2}$.
  3. So $r^2 = 8$ and the equation is $(x-3)^2 + (y-4)^2 = 8$.

Answer: $(x-3)^2 + (y-4)^2 = 8$.

5
Worked Example
Does the equation $x^2 + y^2 + 4x - 6y + 20 = 0$ represent a real circle?
Solution
  1. Identify $g = 2$, $f = -3$, $c = 20$.
  2. Compute $g^2 + f^2 - c = 4 + 9 - 20 = -7$.
  3. Since this is negative, $r^2 < 0$ and no real point satisfies the equation.

Answer: No — it is not a real circle ($g^2+f^2-c = -7 < 0$).

6
Worked Example
Find the equation of the circle with centre $(2, -3)$ that passes through the point $(5, 1)$.
Solution
  1. The radius equals the distance from the centre to the given point.
  2. $r = \sqrt{(5-2)^2 + (1-(-3))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
  3. Equation: $(x-2)^2 + (y+3)^2 = 25$.

Answer: $(x-2)^2 + (y+3)^2 = 25$.

Key Points

  • 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.
Tap an option to check your answer0 / 4
Q1.The equation of a circle with centre origin and radius $r$ is:
Explanation: Standard circle equation.
Q2.The centre of $(x-2)^2+(y+3)^2=16$ is:
Explanation: Centre $(h,k)$ from $(x-h)^2+(y-k)^2$.
Q3.The radius of $x^2+y^2=25$ is:
Explanation: $r=\sqrt{25}=5$.
Q4.For $x^2+y^2+2gx+2fy+c=0$, the centre is:
Explanation: Centre $(-g,-f)$.