Equation of a Circle: Standard and General Forms | NIMCET

Quick answer: A circle with centre $(h, k)$ and radius r is $(x - h)^2 + (y - k)^2 = r^2$. The general form $x^2 + y^2 + 2gx + 2fy + c = 0$ has centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$. The circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$, and the parametric form is $x = h + r\cos\theta$, $y = k + r\sin\theta$.

Forms of the equation

Reading the general form

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

A second-degree equation represents a circle only when the coefficients of $x^2$ and $y^2$ are equal and there is no xy term. If the coefficients are equal but not 1, divide through first.

Special positions

Intercepts: the x-intercept length is $2\sqrt{g^2 - c}$ and the y-intercept length is $2\sqrt{f^2 - c}$.

Position of a point

Let $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.

Position of a line

Let d be the distance from the centre to the line (see the point-to-line distance formula).

The chord length for a secant is $2\sqrt{r^2 - d^2}$.

Worked examples

Example 1: centre and radius

Find the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$.

Step 1: $2g = -6$, $2f = 4$, so $g = -3$, $f = 2$, $c = -12$.

Step 2: Centre $(3, -2)$, radius $\sqrt{9 + 4 + 12} = 5$.

Example 2: diameter form

Find the circle with diameter endpoints $(1, 2)$ and $(5, 6)$.

$(x - 1)(x - 5) + (y - 2)(y - 6) = 0$, so $x^2 + y^2 - 6x - 8y + 17 = 0$.

Check: the centre $(3, 4)$ is the midpoint ✓, and the radius is $\sqrt{9 + 16 - 17} = 2\sqrt{2}$, half of $\sqrt{32}$ ✓.

Example 3: circle through three points

Find the circle through $(0, 0)$, $(4, 0)$ and $(0, 6)$.

Step 1: Through the origin, so $c = 0$.

Step 2: $(4, 0)$: $16 + 8g = 0$, so $g = -2$. $(0, 6)$: $36 + 12f = 0$, so $f = -3$.

Answer: $x^2 + y^2 - 4x - 6y = 0$.

Example 4: chord length

Find the length of the chord cut by $x + y = 1$ on $x^2 + y^2 = 9$.

$d = \frac{1}{\sqrt{2}}$, so the chord length is $2\sqrt{9 - \frac{1}{2}} = 2\sqrt{\frac{17}{2}} = \sqrt{34}$.

Summary table

Exam tips

• Halve the coefficients of x and y before reading the centre.
• A circle through the origin has no constant term.
• For "circle on the line segment as diameter", use the diameter form directly.
• Use $d$ versus $r$ for all line-circle position questions.

Next, study tangents, normals and chord of contact, or see the NIMCET Coordinate Geometry and Vectors chapter.

Common mistakes

• Reading the centre as $(g, f)$ instead of $(-g, -f)$.
• Forgetting to divide by the common coefficient of $x^2$ and $y^2$.
• Treating an equation with an xy term as a circle.
• Adding c instead of subtracting it in the radius formula.

Frequently Asked Questions

What is the general equation of a circle?

x² + y² + 2gx + 2fy + c = 0, with centre (−g, −f) and radius √(g² + f² − c).

How do you find the centre and radius of a circle?

Halve the coefficients of x and y and change their signs to get the centre. The radius is √(g² + f² − c).

What is the diameter form of a circle?

(x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0, where (x₁, y₁) and (x₂, y₂) are the ends of a diameter.

How do you check whether a point lies inside a circle?

Substitute the point into the circle's equation. A negative value means inside, zero means on, and positive means outside.

When is a line a tangent to a circle?

When the perpendicular distance from the centre to the line equals the radius.