In short

From an external point P(x_1, y_1), two tangents can be drawn to a circle S = 0. The pair of tangents has the combined equation SS_1 = T^2. The chord of contact — the line joining the two points of tangency — has the equation T = 0. A chord with a given midpoint M(h, k) has the equation T = S_1 (evaluated at M). Pole and polar extend these ideas to any point, not just external ones.

Stand at a point P outside a circle and stretch two threads from P to the circle so that each thread just touches the circle — no crossing, just grazing. Each thread lies along a tangent line. Where the two threads touch the circle, you get two points. The straight line joining those two contact points is called the chord of contact.

Take P = (7, 1) and the circle x^2 + y^2 = 25. Since S_1 = 49 + 1 - 25 = 25 > 0, the point is outside the circle. Two tangent lines exist. The chord of contact — the line joining the points where those tangents touch the circle — turns out to have the equation 7x + y = 25. That equation looks familiar: it is exactly the T-substitution from the tangent article, applied at the external point rather than at a point on the circle.

This is the key insight of the present article: the T-substitution formula, when applied at different types of points, produces different geometric objects. At a point on the circle, it gives the tangent. At a point outside the circle, it gives the chord of contact. At a point inside, it gives a line called the polar. One formula, three geometric meanings.

Two tangent lines from $P(7, 1)$ to the circle $x^2 + y^2 = 25$. The points of tangency are $A$ and $B$, and the dashed line $AB$ is the chord of contact: $7x + y = 25$.

Equation of the chord of contact

The derivation

Let P(x_1, y_1) be a point outside the circle S \equiv x^2 + y^2 + 2gx + 2fy + c = 0. Suppose the tangent from P touches the circle at the point A(a, b). By the tangent formula from the previous article, the tangent at A is:

xa + yb + g(x + a) + f(y + b) + c = 0

Since this tangent passes through P(x_1, y_1), substituting (x_1, y_1) into the tangent equation gives:

x_1 a + y_1 b + g(x_1 + a) + f(y_1 + b) + c = 0

Rearrange this:

a \cdot x_1 + b \cdot y_1 + g(a + x_1) + f(b + y_1) + c = 0

Now look at this expression. It says: the point A(a, b) satisfies the equation

x \cdot x_1 + y \cdot y_1 + g(x + x_1) + f(y + y_1) + c = 0

where (a, b) has been replaced by the variable point (x, y). By the same argument, if the second tangent touches the circle at B(\alpha, \beta), then B also satisfies this equation. Since both A and B lie on this line, and a line is determined by two points, the equation of the line AB is:

Chord of contact

The chord of contact of the tangents drawn from P(x_1, y_1) to the circle S = 0 is

T = 0

where T \equiv xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c.

For the standard circle x^2 + y^2 = r^2, this simplifies to xx_1 + yy_1 = r^2.

The notation T stands for the expression you get by applying the T-substitution to the circle equation at the point (x_1, y_1). The same letter T appeared in the tangent formula — deliberately, because when (x_1, y_1) is on the circle, T = 0 is the tangent, and when it is outside, T = 0 is the chord of contact.

Verifying with coordinates

Take P(7, 1) and x^2 + y^2 = 25. The chord of contact is T = 0:

7x + y = 25

Does the point A(3, 4) lie on this line? Check: 21 + 4 = 25. Yes. Is A on the circle? 9 + 16 = 25. Yes. So (3, 4) is a point of tangency, and it lies on the chord of contact.

For the second point B, solve 7x + y = 25 and x^2 + y^2 = 25 simultaneously. From the line: y = 25 - 7x. Substitute:

x^2 + (25 - 7x)^2 = 25
x^2 + 625 - 350x + 49x^2 = 25
50x^2 - 350x + 600 = 0
x^2 - 7x + 12 = 0
(x - 3)(x - 4) = 0

So x = 3 (giving A(3, 4)) or x = 4 (giving B(4, -3)). Check B: 16 + 9 = 25. On the circle. And the tangent at B is 4x - 3y = 25. Does P(7, 1) lie on this tangent? 28 - 3 = 25. Yes. Everything is consistent.

Equation of the pair of tangents

The two tangent lines from P to the circle, taken together, form a pair of lines. You can write their combined equation as a single second-degree equation.

The derivation

Let P(x_1, y_1) be external to S \equiv x^2 + y^2 + 2gx + 2fy + c = 0. A general point Q(x, y) lies on one of the two tangent lines from P if and only if the line PQ is tangent to the circle.

The condition for a line to be tangent to a circle is that the perpendicular distance from the centre equals the radius. But there is a cleaner algebraic approach: the combined equation of both tangent lines is

Pair of tangents

The combined equation of the pair of tangents drawn from P(x_1, y_1) to S = 0 is

SS_1 = T^2

where S = x^2 + y^2 + 2gx + 2fy + c, S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c, and T = xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c.

Why does this work? Consider a point Q(x, y) on one of the tangent lines from P. Every point on a tangent line from P to the circle can be parametrised as a point on the line through P and the point of tangency. The expression SS_1 - T^2 is a homogeneous second-degree expression in the deviations (x - x_1) and (y - y_1), and it vanishes precisely when Q lies on one of the two tangent lines. This can be verified directly by expanding:

S \cdot S_1 is the product of the circle's equation evaluated at (x, y) and at (x_1, y_1). T^2 is the square of the T-substitution. When you expand SS_1 - T^2, the result factors into two linear factors — the two tangent lines. The identity SS_1 = T^2 is the condition that the point (x, y) lies on one of those two lines.

A concrete check

For P(7, 1) and x^2 + y^2 = 25:

The pair of tangents is SS_1 = T^2:

25(x^2 + y^2 - 25) = (7x + y - 25)^2

Expand the right side: 49x^2 + 14xy + y^2 - 350x - 50y + 625.

Expand the left side: 25x^2 + 25y^2 - 625.

25x^2 + 25y^2 - 625 = 49x^2 + 14xy + y^2 - 350x - 50y + 625
0 = 24x^2 + 14xy - 24y^2 - 350x - 50y + 1250

Divide by 2: 12x^2 + 7xy - 12y^2 - 175x - 25y + 625 = 0.

This should factor into the two tangent lines. The tangent at A(3, 4) is 3x + 4y - 25 = 0 and the tangent at B(4, -3) is 4x - 3y - 25 = 0. Their product is:

(3x + 4y - 25)(4x - 3y - 25) = 12x^2 - 9xy - 75x + 16xy - 12y^2 - 100y - 100x + 75y + 625
= 12x^2 + 7xy - 12y^2 - 175x - 25y + 625

Confirmed. The formula SS_1 = T^2 exactly produces the combined equation of the two tangent lines.

Chord with a given midpoint

Sometimes the problem runs the other way: you know the midpoint of a chord, and you want the chord's equation.

Let M(h, k) be the midpoint of a chord of the circle S = 0. The centre of the circle is C(-g, -f). The line CM is perpendicular to the chord (a standard property — the line from the centre to the midpoint of a chord is perpendicular to the chord).

The slope of CM is \dfrac{k + f}{h + g}, so the chord has slope -\dfrac{h + g}{k + f}.

Using point-slope form through M(h, k):

y - k = -\frac{h + g}{k + f}(x - h)
(y - k)(k + f) = -(h + g)(x - h)
(h + g)(x - h) + (k + f)(y - k) = 0

Expand: hx + gx - h^2 - gh + ky + fy - k^2 - fk = 0

Add and subtract strategically to match the T-substitution:

xh + yk + g(x + h) + f(y + k) + c = h^2 + k^2 + 2gh + 2fk + c

The left side is T (evaluated at the midpoint (h, k)), and the right side is S_1 (the circle equation evaluated at (h, k)).

Chord with given midpoint

The equation of the chord of S = 0 whose midpoint is M(h, k) is

T = S_1

where T and S_1 are both evaluated at (h, k).

This is elegant: T = 0 is the chord of contact, T = S_1 is the chord with midpoint M, and both are one-line formulas.

A chord of $x^2 + y^2 = 25$ with midpoint $M(3, 1)$. The line from the centre $O$ to $M$ is perpendicular to the chord. The chord equation is $3x + y = 10$ (from $T = S_1$: $3x + y - 25 = 9 + 1 - 25$, i.e. $3x + y = 10$).

Pole and polar

There is a beautiful duality hidden in everything you have seen so far. The T-substitution at a point P always produces a line, regardless of where P is — inside, on, or outside the circle. That line is called the polar of P with respect to the circle, and P is called the pole of that line.

Pole and polar

Given the circle S = 0 and a point P(x_1, y_1), the polar of P with respect to S is the line

T = 0

i.e. xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0.

The point P is called the pole of this line.

The polar unifies three cases:

Position of P S_1 The polar is...
On the circle S_1 = 0 The tangent at P
Outside the circle S_1 > 0 The chord of contact from P
Inside the circle S_1 < 0 A line that does not intersect the circle
Three poles and their polars for $x^2 + y^2 = 16$. $P_1(1, 2)$ is inside the circle — its polar (dashed) does not cross the circle. $P_2(3, 4)$ is on the circle — its polar (solid red) is the tangent. $P_3(4, 0)$ is on the circle — its polar is $x = 4$ (dotted). The same formula $T = 0$ produces all three lines.

The conjugate property

Pole and polar have a remarkable symmetry: if the polar of P passes through Q, then the polar of Q passes through P. This is called the conjugate property.

The proof takes two lines. The polar of P(x_1, y_1) passes through Q(x_2, y_2) when:

x_2 x_1 + y_2 y_1 + g(x_2 + x_1) + f(y_2 + y_1) + c = 0

But this expression is completely symmetric in (x_1, y_1) and (x_2, y_2) — swapping the two points does not change it. So the polar of Q passes through P too. The symmetry of the algebra guarantees the symmetry of the geometry.

Finding the pole of a given line

If you are given the line lx + my + n = 0 and the circle x^2 + y^2 = r^2, and asked to find its pole, compare the polar equation xx_1 + yy_1 = r^2 with lx + my + n = 0. The coefficients must be proportional:

\frac{x_1}{l} = \frac{y_1}{m} = \frac{r^2}{-n}

So the pole is \left(\dfrac{-lr^2}{n}, \dfrac{-mr^2}{n}\right).

Worked examples

Example 1: Chord of contact and tangent pair from a specific point

From the point P(6, 8), find the chord of contact and the length of the chord of contact for the circle x^2 + y^2 = 25.

Step 1. Verify P is external. S_1 = 36 + 64 - 25 = 75 > 0. Yes, P is outside the circle.

Why: the sign of S_1 determines whether P is inside, on, or outside the circle. A positive value means external, and tangent lines exist.

Step 2. Write the chord of contact. T = 0:

6x + 8y = 25

Why: apply the T-substitution at (6, 8) in x^2 + y^2 = 25. Replace x^2 \to 6x, y^2 \to 8y.

Step 3. Find the points of tangency. Solve 6x + 8y = 25 with x^2 + y^2 = 25 simultaneously.

From the line: x = \dfrac{25 - 8y}{6}. Substitute:

\frac{(25 - 8y)^2}{36} + y^2 = 25
625 - 400y + 64y^2 + 36y^2 = 900
100y^2 - 400y - 275 = 0
4y^2 - 16y - 11 = 0

By the quadratic formula: y = \dfrac{16 \pm \sqrt{256 + 176}}{8} = \dfrac{16 \pm \sqrt{432}}{8} = \dfrac{16 \pm 12\sqrt{3}}{8} = \dfrac{4 \pm 3\sqrt{3}}{2}.

Why: solving the chord of contact with the circle finds the exact points of tangency. The quadratic has two roots — one for each tangent point.

Step 4. Compute the length of the chord of contact. The distance from the centre (0, 0) to the chord 6x + 8y - 25 = 0 is:

d = \frac{|0 + 0 - 25|}{\sqrt{36 + 64}} = \frac{25}{10} = 2.5

The half-chord length is \sqrt{r^2 - d^2} = \sqrt{25 - 6.25} = \sqrt{18.75}. So the full chord length is 2\sqrt{18.75} = 5\sqrt{3}.

Result: Chord of contact: 6x + 8y = 25. Length of chord: 5\sqrt{3}.

The circle $x^2 + y^2 = 25$ with the external point $P(6, 8)$ and the chord of contact $6x + 8y = 25$ (dashed). The chord of contact is perpendicular to the line $OP$ — a fact that follows from the symmetry of the two tangent lines.

Notice that P = (6, 8) lies on the line OP with slope 8/6 = 4/3, and the chord of contact 6x + 8y = 25 has slope -6/8 = -3/4. The product of slopes is -1, so OP is perpendicular to the chord of contact. This is always true: the chord of contact is perpendicular to the line joining the external point to the centre.

Example 2: Chord with a given midpoint

Find the equation of the chord of x^2 + y^2 - 4x - 6y + 8 = 0 whose midpoint is M(3, 2).

Step 1. Identify the circle's centre and radius. Here g = -2, f = -3, c = 8, so centre = (2, 3) and R = \sqrt{4 + 9 - 8} = \sqrt{5}.

Why: knowing the centre lets you verify that M is inside the circle (otherwise no chord with midpoint M exists).

Step 2. Check that M(3, 2) is inside the circle. S_1 = 9 + 4 - 12 - 12 + 8 = -3 < 0. Yes, the point is inside.

Step 3. Compute T at (3, 2). Apply the T-substitution:

T = x(3) + y(2) + (-2)(x + 3) + (-3)(y + 2) + 8
= 3x + 2y - 2x - 6 - 3y - 6 + 8 = x - y - 4

And S_1 = -3 (computed in Step 2).

Why: the chord equation is T = S_1, which needs both quantities evaluated at the midpoint.

Step 4. Write the chord equation T = S_1:

x - y - 4 = -3
x - y = 1

Step 5. Verify. The slope of CM (from (2, 3) to (3, 2)) is \dfrac{2 - 3}{3 - 2} = -1. The chord x - y = 1 has slope 1. Since (-1)(1) = -1, the chord is perpendicular to CM. This confirms the answer — the line from the centre to the midpoint of a chord is always perpendicular to the chord.

Result: The chord is x - y = 1.

The circle $(x-2)^2 + (y-3)^2 = 5$ with the chord $x - y = 1$ (red) whose midpoint is $M(3, 2)$. The segment $CM$ is perpendicular to the chord, confirming the result.

The chord x - y = 1 passes through (3, 2), which is the midpoint, and is perpendicular to the line from the centre (2, 3) to that midpoint. The formula T = S_1 encoded both the midpoint condition and the perpendicularity condition in a single equation.

Common confusions

Going deeper

If you came here to learn the chord of contact, chord with midpoint, and pair of tangents formulas, you have everything you need — you can stop here. What follows is for readers interested in the geometric depth of pole and polar, and the elegant harmonic relationship that underpins it.

Pole-polar and harmonic conjugates

The pole-polar relationship has a deeper geometric characterisation involving harmonic division. Given a point P and a circle, draw any line through P that intersects the circle at two points A and B. Let Q be the point on line AB such that P and Q divide AB harmonically — meaning:

\frac{PA}{PB} = \frac{QA}{QB}

(with appropriate signs for directed ratios). The remarkable fact is: no matter which line through P you draw, the point Q always lies on the same line — and that line is the polar of P.

This gives an entirely geometric definition of the polar, independent of coordinates. The polar of P is the locus of the harmonic conjugates of P with respect to all chords of the circle through P.

The pole $P(1, 2)$ and its polar $x + 2y = 16$ (dashed) for $x^2 + y^2 = 16$. Any line through $P$ that cuts the circle at two points $A$, $B$ will meet the polar at the harmonic conjugate of $P$ with respect to $A$ and $B$.

The pole-polar duality as a transformation

The map that sends each point to its polar (and each line to its pole) is an example of a projective transformation called a polarity. It has properties that seem almost magical:

  1. Incidence-preserving: if a point lies on a line, the pole of that line lies on the polar of that point.
  2. Involutory: applying it twice returns to the original configuration (the pole of the polar of P is P itself).
  3. Collinear points map to concurrent lines: if three points are collinear, their polars are concurrent — and vice versa.

These properties are the foundation of projective geometry, which treats poles and polars as the fundamental duality between points and lines. The same structure appears in conics — every ellipse, parabola, and hyperbola defines its own pole-polar duality, with the same algebraic form (T = 0) and the same conjugacy property.

Connection to power of a point

The quantity S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c that appears throughout this article is the power of the point P with respect to the circle. If a line through P intersects the circle at points A and B, then PA \cdot PB = |S_1| (with appropriate signs). This is a constant — it does not depend on which line through P you choose. The power is positive when P is outside (and equals the square of the tangent length), zero when P is on the circle, and negative when P is inside. You will meet the power of a point again in the article on radical axis, where it becomes the key tool for studying families of circles.

Where this leads next

The chord of contact and pole-polar are building blocks for more advanced circle geometry: