Circles: Tangents, Points of Contact, and Equal Lengths
Two clean theorems power this whole chapter: a tangent is always perpendicular to the radius at its point of contact, and the two tangents drawn from any external point are always equal in length. This companion covers both proofs, the worked examples that build on them, and a CBSE-style quiz to test yourself.
tangent ⊥ radiusPQ = PRpoint of contactone line, one touch
⭕10.1 Lines and Circles: Three Possibilities
You already know a circle is the collection of all points in a plane at a constant distance (the radius) from a fixed point (the centre), and you've met terms like chord, segment, sector, and arc. This chapter asks a new question: when you draw a straight line and a circle together on the same plane, what can actually happen between them?
Non-intersecting
The line and circle share no common point.
Secant
The line and circle share exactly two common points, A and B.
Tangent
The line and circle share exactly one common point, A.
↔️Non-intersecting Line. A line that shares no common point with a circle is called a non-intersecting line with respect to that circle.
✂️Secant. A line that intersects a circle at exactly two points is called a secant of the circle.
👆Tangent. A line that intersects a circle at exactly one point is called a tangent to the circle.
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You've already seen one in real life.. Think of a pulley over a well: the rope on either side, treated as a straight line, touches the wheel of the pulley at exactly one point — behaving exactly like a tangent to the circle representing the pulley.
These three cases — non-intersecting, secant, and tangent — are the ONLY possible relationships between a line and a circle in a plane; there's no fourth option. The rest of this chapter studies tangents specifically: why they exist, and what special properties they have.
📍10.2 Tangent to a Circle
Picture a straight wire pinned at a point P on a circular wire, free to rotate around P. As you rotate it, most positions cut the circle at P and at some second point. But at exactly one special position, the straight wire touches the circle only at P — that position is the tangent at P.
🎯One and only one tangent at each point.. As the rotating wire approaches that special tangent position, the second intersection point slides closer and closer to P, and coincides with it exactly at the tangent position. Rotate further past it, and a new second intersection point appears on the other side — there is only ever one tangent line at any given point of a circle.
A closely related activity: draw a circle and a secant, then draw more lines parallel to that secant on both sides. As each parallel line moves outward, the chord it cuts gets shorter and shorter, shrinking toward zero length right as the line becomes tangent. This shows something useful: a tangent is simply a secant in the special case where the two endpoints of its chord have coincided into a single point — and it also shows that a circle can have at most two tangents parallel to any given secant, one on each side.
The single shared point between a tangent and a circle is called the point of contact, and the tangent is said to touch the circle there.
🚲
Look at a rolling bicycle wheel.. All the spokes of a wheel lie along its radii. As the wheel rolls, the ground it moves along behaves exactly like a tangent line to the circle of the wheel — and at every instant, the spoke touching the ground looks like it meets the ground at a perfect right angle.
📐Theorem 10.1
📐Theorem 10.1. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: a circle with centre O, and a tangent XY touching it at point P. We need to prove OP ⊥ XY.
PICK ANY OTHER POINT ON THE TANGENTTake any point Q on XY other than P, and join OQ.
Q MUST LIE OUTSIDE THE CIRCLEIf Q were inside the circle, the line XY would cross the circle at a second point too, making it a secant rather than a tangent — contradiction. So Q lies outside.
COMPARE DISTANCESSince Q is outside the circle while P is on it, OQ is longer than the radius OP. So OQ > OP for every point Q on XY other than P.
CONCLUDESince OP is shorter than the distance from O to every other point on line XY, OP is the shortest distance from O to the line — and the shortest distance from a point to a line is always along the perpendicular. So OP ⊥ XY.
📌Two useful remarks.. (1) Since this proof holds for any tangent, there is exactly one tangent possible at any given point on a circle. (2) The line containing the radius through the point of contact is sometimes also called the 'normal' to the circle at that point.
📏10.3 Number of Tangents from a Point
Instead of asking how many tangents pass through a point ON the circle, let's ask: how many tangents can be drawn from a point in relation to the circle, depending on where that point sits?
Point inside the circle
Every line through it is a secant — no tangent is possible.
Point on the circle
Exactly one tangent exists, as shown by Theorem 10.1.
Point outside the circle
Exactly two tangents can be drawn, touching at two distinct points.
Case 1: There is NO tangent to a circle through a point lying inside it.
Case 2: There is exactly ONE tangent to a circle through a point lying on it.
Case 3: There are exactly TWO tangents to a circle through a point lying outside it.
For a point P outside the circle, with tangents touching at points T₁ and T₂, the segment from P to either point of contact is called the length of the tangent from P. A natural question follows: do PT₁ and PT₂ always come out equal?
📐Theorem 10.2
📐Theorem 10.2. The lengths of the two tangents drawn from an external point to a circle are equal.
Given: a circle with centre O, an external point P, and two tangents PQ and PR touching the circle at Q and R. We need to prove PQ = PR.
JOIN THE RADIIJoin OP, OQ, and OR.
IDENTIFY TWO RIGHT ANGLESBy Theorem 10.1, the tangent at each point of contact is perpendicular to the radius there, so ∠OQP = ∠ORP = 90°.
COMPARE THE TWO RIGHT TRIANGLESIn right triangles OQP and ORP: OQ = OR (radii of the same circle), and OP = OP (shared side).
CONCLUDE VIA RHS CONGRUENCE△OQP ≅ △ORP (RHS), so PQ = PR (CPCT).
📌Two more useful remarks.. (1) You could equally prove this with Pythagoras: PQ² = OP²−OQ² = OP²−OR² = PR², since OQ=OR. (2) Since the triangles are congruent, ∠OPQ = ∠OPR too — so OP bisects the angle between the two tangents, meaning the centre always lies on the bisector of the angle formed by two tangents from any external point.
✏️Worked Examples
Prove that in two concentric circles, a chord of the larger circle that touches the smaller circle is bisected at the point of contact.
SET UPLet AB be a chord of the larger circle C₁ (centre O) that touches the smaller circle C₂ at point P. Join OP.
APPLY THEOREM 10.1Since AB is a tangent to C₂ at P with radius OP, Theorem 10.1 gives OP ⊥ AB.
USE THE PERPENDICULAR-BISECTS-CHORD FACTAB is also a chord of the larger circle C₁, and the perpendicular from a circle's centre to any chord always bisects that chord. So AP = BP.
📝10.4 Chapter Summary
A tangent to a circle is a line that intersects the circle at exactly one point — a special limiting case of a secant, where the chord's two endpoints have merged into one.
Theorem 10.1: the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Theorem 10.2: the lengths of the two tangents drawn from an external point to a circle are always equal — and the segment from the centre to that external point always bisects the angle between the two tangents.
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Two theorems, endless applications.. These two facts — perpendicularity at the point of contact, and equal tangent lengths — are the backbone of countless circle-geometry proofs involving circumscribed quadrilaterals, triangles with incircles, and more, which you'll build on in later problems.
TOOL 1 Tangent Length Calculator
Enter any two of the three quantities — the circle's radius, the distance from an external point to the centre, and the tangent length — and see the third solved via the Pythagoras relationship from Theorem 10.2, with every step shown.
IDENTIFY THE RIGHT TRIANGLE
Theorem 10.1: the radius is ⊥ the tangent at the point of contact, so radius, tangent, and the segment to the external point form a right triangle with the external-point segment as hypotenuse.
APPLY PYTHAGORAS
radius² = distance² − tangent² = 25² − 24² = 49
TAKE THE SQUARE ROOT
radius = √49 = 7
Radius: 7
Distance to centre: 25
Tangent length: 24
TOOL 2 Chord & Tangent Intersection Solver
Enter a chord's length and the circle's radius, and see the length of the tangent segment (from the point where the two tangents at the chord's endpoints meet) worked out step by step, exactly as in Example 3.
SET UP
OT bisects the chord PQ, so PR = RQ = 8/2 = 4 cm.
FIND OR
In right triangle OPR: OR = √(OP² − PR²) = √(5² − 4²) = 3 cm.
SPOT A SIMILAR TRIANGLE
Right triangle TRP is similar to right triangle PRO (AA similarity), since both share the angle complementary to ∠RPO.
SOLVE THE PROPORTION
TP/PO = RP/RO ⟹ TP = 5 × 4 / 3 = 6.6667 cm
Half-chord (PR): 4
Perpendicular distance (OR): 3
Tangent length (TP): 6.6667
§3 Flashcards
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Card 1 / 12
Q
What are the three possible relationships between a line and a circle?
A
Non-intersecting (no common point), secant (two common points), and tangent (exactly one common point).
§4 Quiz
Modelled on the CBSE Class 10 Section A paper — Multiple Choice and Assertion–Reason questions, each worth 1 mark. 20 questions per attempt, drawn from a pool of 40.
