Audited 05 Aug 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Tangent Circle Calculator

Length of the tangent from an outside point to a circle, from the distance to the centre and the radius: t = √(d² − r²), the right-triangle rule.

Tangent Circle Calculator

Tangent segment length
12
The distance from the outside point to where either tangent touches the circle, in the same unit as your inputs.

Background.

From a point outside a circle you can draw exactly two straight lines that just graze it, one on each side. Each touches the circle at a single point, and the stretch of line from your point to that touch point is the tangent length. The two are always equal, which is why a cord pulled taut from a peg around both sides of a round post leaves the post at matching distances.

The length follows from one fact. A radius drawn to the point of contact is perpendicular to the tangent there. So the circle's centre, the point of contact and your outside point form a right-angled triangle, with the line from the centre to your point as its hypotenuse. Pythagoras finishes the job: the tangent length squared equals the distance to the centre squared, minus the radius squared.

The best-known use of that triangle is the horizon. Your eye is the outside point, the Earth is the circle, and the tangent length is how far you can see across a calm sea before the surface curves away. With the Earth's mean radius of 6,371 km and eyes 1.7 m above the water, the geometric horizon is about 4.65 km off, and atmospheric refraction stretches the visible horizon a little beyond that.

You supply two numbers in the same unit: the distance from the point to the centre of the circle, and the radius. The answer comes back in that unit. The point has to lie outside the circle, so the distance must be larger than the radius; a point on the circle has a tangent length of zero, and from a point inside, every line cuts the circle twice.

What is tangent circle calculator?

The tangent length is the distance along a tangent line from an external point to the point where that line touches the circle. Every external point has two such tangents, mirror images of each other across the line to the centre, and their lengths are equal. Squared, the tangent length is also the power of the point with respect to the circle, d² − r², which equals the product of the two distances along any secant drawn from the point through the circle.

How to use this calculator.

  1. Work out the distance from the outside point to the circle's centre, not to its edge. If you know the gap to the nearest point of the circle instead, add the radius: a point 8 units from the edge of a radius-5 circle is 13 from the centre.
  2. Enter the radius in the same unit. If you were given a diameter, halve it first.
  3. Keep the distance larger than the radius. A value equal to or below it is refused, because the point would sit on the circle or inside it.
  4. Read the tangent length in the unit you used. It is the length of each of the two tangents from the point.

The formula.

Tangent length = √(d²−r²)

t = √(d² − r²). Here d is the distance from the outside point P to the centre O, r is the radius and t is the distance from P to the point of contact T. The radius OT meets the tangent at T at a right angle; if it did not, the line would dip into the circle instead of touching it. Triangle OTP therefore has its right angle at T and its hypotenuse along OP, so Pythagoras gives d² = r² + t², and rearranging gives the formula. Requiring d to exceed r keeps the quantity under the square root positive. As P moves away, t approaches d, because the radius becomes a small part of the triangle; as P closes in on the circle, t shrinks quickly toward zero. The same quantity d² − r² equals PA × PB for any line from P that cuts the circle at A and B.

A worked example.

Example

A point sits 13 units from the centre of a circle of radius 5. Squaring both gives 13² = 169 and 5² = 25. The difference is 169 − 25 = 144, and √144 = 12, so the tangent length is 12. This is the 5-12-13 right triangle: the radius of 5 and the tangent of 12 are the two legs, and the 13-unit line to the centre is the hypotenuse, with the right angle at the point of contact. Double the distance to 26 and the tangent becomes √(676 − 25) = √651 ≈ 25.515, now almost as long as the line to the centre. Bring the point in to 5.5 and it falls to √5.25 ≈ 2.291.

center Distance13
radius5

Frequently asked questions.

Why are the two tangents from one point always equal?
The two right triangles they form share the hypotenuse from the point to the centre and have equal legs, since both radii are r. By Pythagoras their tangent legs must match. The same fact is why a triangle's inscribed circle touches the two sides meeting at any corner at equal distances from that corner.
How far away is the horizon from a given height?
Treat the Earth as the circle. In kilometres, enter the radius 6,371 and a distance of 6,371 plus your eye height. From 1.7 m (6,371.0017 km) the tangent length is about 4.654 km; from a 100 m cliff (6,371.1 km) it is about 35.7 km. Refraction lets you see slightly farther than this geometric figure.
How do I find the angle between the two tangents?
The page does not output it, but it follows from the same triangle: half the angle at the outside point has a sine of r ÷ d. For a point 13 from the centre of a radius-5 circle, that half-angle is about 22.62°, so the two tangents meet at about 45.24°.
What does the power of a point have to do with tangent length?
The power of a point is d² − r², the tangent length squared. It also equals PA × PB for any line from the point that crosses the circle at A and B. A line from the point 13 units out straight through the centre meets the radius-5 circle at 8 and 18 units, and 8 × 18 = 144 = 12².
Can I use this for a tangent shared by two circles?
Indirectly. For two circles whose centres are D apart, the outer common tangent has length √(D² − (r₁ − r₂)²) and the crossing inner tangent √(D² − (r₁ + r₂)²). Enter D as the distance and r₁ − r₂ or r₁ + r₂ as the radius. With D = 10, r₁ = 5 and r₂ = 2, those give about 9.539 and 7.141.

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Tangent length = √(d²−r²)
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