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.
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:
Since this tangent passes through P(x_1, y_1), substituting (x_1, y_1) into the tangent equation gives:
Rearrange this:
Now look at this expression. It says: the point A(a, b) satisfies the equation
where (a, b) has been replaced by the variable point (x, y).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
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:
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:
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
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:
- S = x^2 + y^2 - 25
- S_1 = 49 + 1 - 25 = 25
- T = 7x + y - 25
The pair of tangents is SS_1 = T^2:
Expand the right side: 49x^2 + 14xy + y^2 - 350x - 50y + 625.
Expand the left side: 25x^2 + 25y^2 - 625.
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:
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):
Expand: hx + gx - h^2 - gh + ky + fy - k^2 - fk = 0
Add and subtract strategically to match the T-substitution:
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
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.
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
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 |
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:
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:
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:
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:
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:
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}.
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:
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:
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 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
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"The chord of contact is a tangent." No. The chord of contact is a chord — it has both endpoints on the circle. It is the line joining the two points where the tangents from an external point touch the circle. The tangents themselves are different lines.
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"T = 0 at an external point gives the tangent." It gives the chord of contact, not the tangent. The tangent formula T = 0 applies when the point is on the circle. When the point is external, T = 0 gives the chord of contact. Same formula, different geometric meaning depending on where the point sits.
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"The formula SS_1 = T^2 gives a circle." No — it gives a pair of straight lines. The equation is second-degree, but it represents two lines (the two tangent lines), not a conic curve. You can verify this by checking that the discriminant of the conic is zero, which is the condition for a degenerate conic (pair of lines).
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"Pole and polar only exist for external points." Every point in the plane has a polar, and every line has a pole — regardless of position relative to the circle. The polar of an internal point is a line that does not intersect the circle. The polar of the centre is the "line at infinity," which is why the centre has no finite polar.
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:
(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-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:
- Incidence-preserving: if a point lies on a line, the pole of that line lies on the polar of that point.
- Involutory: applying it twice returns to the original configuration (the pole of the polar of P is P itself).
- 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:
- Tangent and Normal to Circle — the prerequisite to this article, covering the tangent at a point and the tangent with given slope.
- Family of Circles — circles through the intersection of two circles, the radical axis (which is secretly a chord-of-contact idea), and orthogonal circles.
- Circle — Standard Forms — the different ways to write the equation of a circle.
- Line and Circle — the intersection theory that underpins tangency conditions.
- Conic Sections — Introduction — the same pole-polar duality extends to parabolas, ellipses, and hyperbolas.