Circle — Special Cases

Take three pins and push them into a sheet of paper — anywhere, as long as they're not in a straight line. Now try to draw a circle that passes through all three pins. With a steady hand and some patience, you will find that exactly one circle does the job. Not two, not zero — one.

That is a remarkable fact. Two points don't determine a circle — infinitely many circles pass through any two given points. But the moment you fix a third point (not on the line through the first two), the circle snaps into place. The geometric reason is that the centre must be equidistant from all three points, and as you will see, that pins down the centre to a single location.

This article covers four constructions that every coordinate-geometry course expects you to handle: the circle through three points, the diameter form, parametric equations, and the position of a point relative to a circle.

Circle through three given points

The method

Suppose you want the circle through three points A = (x_1, y_1), B = (x_2, y_2), C = (x_3, y_3). Since every circle has the general form x^2 + y^2 + 2gx + 2fy + c = 0, and the three points must satisfy this equation, you get three linear equations in the three unknowns g, f, c:

Why: each equation says "this point lies on the circle." The terms x_i^2 + y_i^2 are known numbers (computed from the given coordinates), so these are genuinely three linear equations in g, f, c.

Solve this 3 \times 3 system — by substitution, elimination, or Cramer's rule — and you have the circle's equation. The system has a unique solution exactly when the three points are non-collinear.

Why three non-collinear points work

The centre of the circle must be equidistant from all three points. The locus of points equidistant from A and B is the perpendicular bisector of AB. The locus of points equidistant from B and C is the perpendicular bisector of BC. The centre is the intersection of these two perpendicular bisectors.

Two distinct lines in the plane either intersect at exactly one point (if they are not parallel) or not at all (if they are parallel). The perpendicular bisectors of AB and BC are parallel only when A, B, C are collinear — because in that case, both bisectors are perpendicular to the same line, hence parallel to each other. If the three points are not collinear, the two bisectors meet at exactly one point: the centre.

A full computation

Find the circle through A = (0, 0), B = (8, 0), C = (0, 6).

Substituting into x^2 + y^2 + 2gx + 2fy + c = 0:

From A(0, 0):

From B(8, 0):

From C(0, 6):

Why: having one point at the origin simplifies things enormously — it immediately gives c = 0, and the other two equations each contain only one unknown.

The equation is x^2 + y^2 - 8x - 6y = 0. Centre = (-g, -f) = (4, 3), radius = \sqrt{16 + 9 - 0} = 5.

Verify: A(0,0) gives 0 + 0 - 0 - 0 = 0. B(8,0) gives 64 + 0 - 64 - 0 = 0. C(0,6) gives 0 + 36 - 0 - 36 = 0. All three check out.

Notice something about the triangle ABC: its sides are AB = 8, AC = 6, and BC = \sqrt{64 + 36} = 10. Since 6^2 + 8^2 = 10^2, this is a right triangle with the right angle at A. The hypotenuse BC has length 10, and the radius of the circle is 5 = 10/2. This is not a coincidence — the circle through the vertices of a right triangle always has the hypotenuse as a diameter. You will see why in the next section.

The diameter form

When you know that two points A = (x_1, y_1) and B = (x_2, y_2) are endpoints of a diameter of a circle, there is a shortcut that gives the equation without computing the centre and radius separately.

Derivation

Let P = (x, y) be any point on the circle (other than A or B). Since AB is a diameter, the angle \angle APB is a right angle — this is the famous angle-in-a-semicircle theorem (Thales' theorem).

The slope of PA is \dfrac{y - y_1}{x - x_1} and the slope of PB is \dfrac{y - y_2}{x - x_2}.

Two lines are perpendicular when the product of their slopes is -1:

Multiply both sides by (x - x_1)(x - x_2):

Rearrange:

Why: the angle-in-a-semicircle condition (perpendicularity of PA and PB) translates directly into the slope condition, which simplifies to this clean equation. The points A and B themselves also satisfy this equation (each makes one factor zero), so the equation captures the entire circle.

This is elegant: no need to find the midpoint, no need to compute the radius, no square roots at all. The diameter endpoints go directly into the equation.

Expand it and you get the general form. The centre is \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) — the midpoint of the diameter, as expected — and the radius is \dfrac{1}{2}\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} — half the length of the diameter.

Parametric equations of a circle

The idea

In the standard form (x - h)^2 + (y - k)^2 = r^2, the point (x, y) is constrained by a relationship between x and y. But there is another way to describe the same circle: let a single parameter — an angle \theta — trace out every point as it sweeps from 0 to 2\pi.

Derivation

Consider the circle x^2 + y^2 = r^2 (centred at the origin, for simplicity). A point P on this circle is at distance r from the origin. If the line OP makes an angle \theta with the positive x-axis, then by the definition of \cos and \sin:

Why: in a right triangle with hypotenuse r and angle \theta, the adjacent side is r\cos\theta and the opposite side is r\sin\theta. This is simply the definition of trigonometric ratios applied to the radius.

Verify: (r\cos\theta)^2 + (r\sin\theta)^2 = r^2(\cos^2\theta + \sin^2\theta) = r^2 \cdot 1 = r^2. The Pythagorean identity \cos^2\theta + \sin^2\theta = 1 guarantees that every point of this form lies on the circle. And conversely, every point on the circle corresponds to some angle \theta.

For a circle with centre (h, k) and radius r, shift the origin:

The parametric form is useful when you want to "walk along" the circle — for instance, to find where a moving point on the circle is closest to a given line, or to set up optimisation problems. Any point on the circle can be written as (h + r\cos\theta, k + r\sin\theta) for some \theta, which reduces a two-variable problem to a single-variable problem in \theta.

Position of a point with respect to a circle

Inside, on, or outside?

Given a circle S: x^2 + y^2 + 2gx + 2fy + c = 0 and a point P = (x_1, y_1), there are three possibilities: P is inside the circle, on the circle, or outside the circle.

The test is simple. Compute

This is the value you get by substituting (x_1, y_1) into the left side of the circle's equation.

• If S_1 = 0: the point is on the circle.
• If S_1 < 0: the point is inside the circle.
• If S_1 > 0: the point is outside the circle.

Why the sign works

Rewrite the circle in standard form: (x + g)^2 + (y + f)^2 = g^2 + f^2 - c. The left side is the squared distance from (x, y) to the centre (-g, -f). Call this d^2, and call the radius r^2 = g^2 + f^2 - c.

Then S_1 = d^2 - r^2, where d is the distance from P to the centre.

• d = r means P is on the circle: S_1 = 0.
• d < r means P is closer to the centre than the radius — inside: S_1 < 0.
• d > r means P is farther from the centre than the radius — outside: S_1 > 0.

Why: (x_1 + g)^2 + (y_1 + f)^2 - (g^2 + f^2 - c) = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c = S_1. So S_1 is literally d^2 - r^2, and the sign of that difference tells you whether d is less than, equal to, or greater than r.

Verify with the circle (x - 2)^2 + (y - 3)^2 = 16, which in general form is x^2 + y^2 - 4x - 6y - 3 = 0.

• P_1 = (3, 4): S_1 = 9 + 16 - 12 - 24 - 3 = -14 < 0. Inside.
• P_2 = (6, 3): S_1 = 36 + 9 - 24 - 18 - 3 = 0. On the circle.
• P_3 = (7, 6): S_1 = 49 + 36 - 28 - 36 - 3 = 18 > 0. Outside.

For P_1: distance from (3, 4) to centre (2, 3) is \sqrt{1 + 1} = \sqrt{2} \approx 1.41, which is less than the radius 4. Inside. For P_3: distance from (7, 6) to (2, 3) is \sqrt{25 + 9} = \sqrt{34} \approx 5.83, which is greater than 4. Outside. The sign test and the distance comparison agree.

Worked examples

Common confusions

• "Any three points determine a circle." Only if they are non-collinear. Three collinear points (all on the same line) do not lie on any circle. When you set up the 3 \times 3 system for g, f, c, you will find it has no solution — the determinant is zero.

• "The diameter form works for any two points on the circle." Only for two points that are endpoints of a diameter — a chord passing through the centre. If the two points are endpoints of an arbitrary chord (not a diameter), the diameter form gives a different, larger circle, not the one you want.

• "\theta in the parametric form is the angle at the point." No — \theta is the angle at the centre, measured from the positive x-direction. The angle subtended at a point on the circle (the inscribed angle) is a different thing entirely.

• "If S_1 > 0, the point is inside." The signs are the opposite of what many students first guess. S_1 > 0 means the point is outside, and S_1 < 0 means inside. A mnemonic: when you substitute the centre itself, you get g^2 + f^2 + 2g(-g) + 2f(-f) + c = -g^2 - f^2 + c = -(g^2 + f^2 - c) = -r^2 < 0. The centre is inside, and the value is negative. So negative means inside.

• "I can use any form of the circle equation for the point-position test." The sign convention S_1 < 0 for "inside" assumes the coefficient of x^2 is +1. If the equation is 2x^2 + 2y^2 + \ldots = 0, divide by 2 first, or the signs will be wrong by a factor of 2.

Going deeper

The circumcircle of a triangle

The circle through three non-collinear points is the circumcircle of the triangle formed by those three points. Its centre is called the circumcentre, and its radius is the circumradius, usually denoted R.

There is a beautiful formula connecting the circumradius to the triangle's sides. If the triangle has sides a, b, c and area \Delta, then

This follows from the extended law of sines: \dfrac{a}{\sin A} = 2R. The circumcentre is equidistant from all three vertices, and it lies at the intersection of the perpendicular bisectors of the sides — the same geometric construction described earlier.

For a right triangle, the circumcentre is the midpoint of the hypotenuse, and the circumradius is half the hypotenuse. This is why the circle through A(0,0), B(8,0), C(0,6) had its centre at (4, 3) — the midpoint of the hypotenuse BC — and radius 5 — half of BC = 10.

Parametric form and trigonometric substitution

The parametric equations x = h + r\cos\theta, y = k + r\sin\theta are useful beyond just describing the circle. They are a standard tool in optimisation problems involving circles.

For instance: find the maximum value of 3x + 4y on the circle x^2 + y^2 = 25. Using the parametric form with h = 0, k = 0, r = 5:

This is a single-variable expression in \theta. Its maximum value is \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25. The parametric form turned a constrained two-variable problem into an unconstrained one-variable problem — a technique you will see again and again in optimisation.

The power of a point

The quantity S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c that you compute for the point-position test has a deeper geometric meaning. It is called the power of the point P with respect to the circle.

If P is outside the circle and you draw any line through P that intersects the circle at points A and B, then PA \cdot PB = S_1. This product is the same regardless of which line you draw — it depends only on P and the circle, not on the particular line. That is why S_1 is called the "power": it measures a fundamental relationship between the point and the circle.

If P is inside the circle, the power is negative, and the absolute value equals PA \cdot PB where A and B are the two points where any chord through P meets the circle. If P is on the circle, the power is zero. The power-of-a-point concept leads to the radical axis (the locus of points having equal power with respect to two circles), which is a straight line — an idea you will meet in more advanced coordinate geometry.

Where this leads next

The constructions in this article — finding circles through given points, using the diameter form, parametric equations, and point position — are the tools you need for the next level of circle geometry.

• Line and Circle — what happens when a straight line meets a circle: intersection, tangency, chord length, and the equation of a tangent in all its forms.
• Tangent and Normal to Circle — the tangent at a point on the circle, tangents from an external point, and the normal line.
• Pair of Tangents and Chord of Contact — the two tangent lines from an external point and the chord connecting their points of tangency.
• Circle — Standard Forms — the foundation article on the standard and general forms, if you want to review.
• Systems of Linear Equations — the theory behind solving the 3 \times 3 system that finds the circle through three points.