Circles

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CLASS X Mathematics ~3 marks/year Ch 10 of 14
Circles

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • A line and a circle meet in 0, 1 or 2 points — giving a non-intersecting line, a tangent (1 point) or a secant (2 points).
  • Theorem 10.1: the tangent at any point is perpendicular to the radius through the point of contact — so radius $\perp$ tangent always.
  • Theorem 10.2: the two tangents drawn from an external point have equal length; the centre lies on the bisector of the angle between them.
  • From an external point there are exactly two tangents; from a point on the circle, exactly one; from a point inside, none.
  • Board weightage: ~3 marks/year — usually one tangent-length calculation (1–2 marks) or one short proof using the two theorems (3 marks).
Detailed notes

1. Where this chapter sits

In Class 9 you learned that a circle is the set of all points in a plane at a fixed distance (the radius) from a fixed point (the centre), together with terms like chord, arc, segment and sector. This chapter asks a new question: what happens when a straight line meets a circle?

The whole chapter rests on just two ideas — definitions of tangent and secant, and two theorems about tangents. It is short but scoring: the proofs are standard and the numerical questions are quick once you see the right-angled triangle hiding inside.

2. A line and a circle — three possibilities

Take a circle and a line $PQ$ in the same plane. Exactly three situations can occur, and no others:

  • Non-intersecting line: the line and circle have no common point. The line stays entirely outside.
  • Secant: the line cuts the circle at two distinct points $A$ and $B$. The chord $AB$ is the piece inside.
  • Tangent: the line touches the circle at exactly one point $A$. That single point is the point of contact, and the line is said to touch the circle there.

The word "tangent" comes from the Latin tangere, "to touch" (introduced by Thomas Fineke in 1583). Everyday tangents: the rope over a well-pulley, and the ground beneath a moving bicycle wheel — the wheel rolls along a line tangent to itself.

3. Tangent as a limiting secant

Imagine a secant cutting the circle at $A$ and $B$. Slide the line so that $B$ creeps towards $A$ along the circle. The chord $AB$ shrinks; when $B$ finally coincides with $A$, the secant has become a tangent. This gives the key picture:

$$\text{Tangent} = \text{secant in which the two end-points of the chord coincide.}$$

Two consequences you should remember:

  • At any given point on a circle there is one and only one tangent.
  • There can be at most two tangents parallel to any given secant (one on each side).

4. Theorem 10.1 — tangent is perpendicular to the radius

Statement: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

NCERT Theorem 10.1 — full proof

Given: a circle with centre $O$ and a tangent $XY$ touching it at point $P$. To prove: $OP \perp XY$.

1. Take any point $Q$ on $XY$ other than $P$, and join $OQ$.

2. Since $P$ is the only point of $XY$ on the circle, every other point of $XY$ — including $Q$ — lies outside the circle. (If $Q$ were inside, $XY$ would cut the circle twice and be a secant, not a tangent.)

3. Therefore $OQ$ is longer than the radius $OP$, i.e. $OQ > OP$.

4. This holds for every point $Q$ on $XY$ except $P$. So $OP$ is the shortest distance from $O$ to the line $XY$.

5. The shortest distance from a point to a line is the perpendicular distance. Hence $OP \perp XY$. $\blacksquare$

$$\text{radius at point of contact} \perp \text{tangent}$$

Remarks: (i) it follows that at any point on a circle there is exactly one tangent; (ii) the line through the radius at the point of contact is called the normal to the circle there.

5. Number of tangents from a point

How many tangents can you draw to a circle through a given point? It depends entirely on where the point is:

  • Point inside the circle: every line through it cuts the circle twice — no tangent is possible.
  • Point on the circle: exactly one tangent (the line $\perp$ to the radius there).
  • Point outside the circle: exactly two tangents.

For an external point $P$, the two points of contact $T_1, T_2$ give two tangent segments $PT_1$ and $PT_2$. The length of the tangent from $P$ is the distance from $P$ to its point of contact. A circle as a whole has infinitely many tangents (one at every point of it).

6. Theorem 10.2 — tangents from an external point are equal

Statement: The lengths of tangents drawn from an external point to a circle are equal.

NCERT Theorem 10.2 — full proof

Given: a circle with centre $O$, an external point $P$, and two tangents $PQ$ and $PR$ with points of contact $Q$ and $R$. To prove: $PQ = PR$.

1. Join $OP$, $OQ$ and $OR$.

2. By Theorem 10.1, $\angle OQP = \angle ORP = 90^\circ$ (radius $\perp$ tangent).

3. In right triangles $OQP$ and $ORP$:

  • $OQ = OR$ (radii of the same circle),
  • $OP = OP$ (common hypotenuse),
  • $\angle OQP = \angle ORP = 90^\circ$.

4. So $\triangle OQP \cong \triangle ORP$ by the RHS congruence rule.

5. Hence $PQ = PR$ by CPCT. $\blacksquare$

Alternative (Pythagoras): $PQ^2 = OP^2 - OQ^2 = OP^2 - OR^2 = PR^2$, so $PQ = PR$.

$$PQ = PR, \qquad \text{tangent length} = \sqrt{OP^2 - r^2}$$

Bonus fact: the same congruence gives $\angle OPQ = \angle OPR$, so $OP$ bisects the angle $\angle QPR$ between the two tangents — the centre lies on that bisector.

7. Worked NCERT Examples

NCERT Example 1 — chord of larger concentric circle

Prove: in two concentric circles, a chord of the larger circle that touches the smaller circle is bisected at the point of contact.

Let $C_1$ (large) and $C_2$ (small) share centre $O$, and let chord $AB$ of $C_1$ touch $C_2$ at $P$. Join $OP$. Now $AB$ is tangent to $C_2$ at $P$ and $OP$ is its radius, so by Theorem 10.1, $OP \perp AB$. But $AB$ is a chord of $C_1$, and the perpendicular from the centre to a chord bisects it. Hence $AP = BP$ — the point of contact bisects the chord. $\blacksquare$

NCERT Example 2 — angle between tangents

Prove: two tangents $TP, TQ$ from external point $T$ satisfy $\angle PTQ = 2\,\angle OPQ$.

Let $\angle PTQ = \theta$. By Theorem 10.2, $TP = TQ$, so $\triangle TPQ$ is isosceles and

$\angle TPQ = \angle TQP = \dfrac{1}{2}(180^\circ - \theta) = 90^\circ - \dfrac{1}{2}\theta.$

By Theorem 10.1, $\angle OPT = 90^\circ$. Therefore

$\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \dfrac{1}{2}\theta\right) = \dfrac{1}{2}\theta = \dfrac{1}{2}\angle PTQ.$

Hence $\angle PTQ = 2\,\angle OPQ$. $\blacksquare$

NCERT Example 3 — find tangent length TP

$PQ$ is a chord of length $8$ cm of a circle of radius $5$ cm. The tangents at $P$ and $Q$ meet at $T$. Find $TP$.

Join $OT$, meeting $PQ$ at $R$. Since $\triangle TPQ$ is isosceles and $TO$ bisects $\angle PTQ$, we get $OT \perp PQ$ and $OT$ bisects $PQ$, so $PR = RQ = 4$ cm.

In right $\triangle OPR$: $OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = \sqrt{9} = 3$ cm.

Now $\angle TPR + \angle RPO = 90^\circ$ and $\angle TPR + \angle PTR = 90^\circ$, so $\angle RPO = \angle PTR$; thus right triangles $TRP$ and $PRO$ are similar (AA). Hence

$\dfrac{TP}{PO} = \dfrac{RP}{RO} \;\Rightarrow\; \dfrac{TP}{5} = \dfrac{4}{3} \;\Rightarrow\; TP = \dfrac{20}{3}$ cm.

8. NCERT Exercise 10.1 — fully solved

Q1. How many tangents can a circle have? A circle has a tangent at every one of its points, so it has infinitely many tangents.

Q2. Fill in the blanks:

  • (i) A tangent to a circle intersects it in one point(s).
  • (ii) A line intersecting a circle in two points is called a secant.
  • (iii) A circle can have two parallel tangents at the most.
  • (iv) The common point of a tangent to a circle and the circle is called the point of contact.

Q3. Tangent $PQ$ at $P$ of a circle of radius $5$ cm meets a line through centre $O$ at $Q$ with $OQ = 12$ cm. Since $OP \perp PQ$ (Theorem 10.1), $\triangle OPQ$ is right-angled at $P$:

$PQ = \sqrt{OQ^2 - OP^2} = \sqrt{12^2 - 5^2} = \sqrt{144 - 25} = \sqrt{119}$ cm. Answer: (D) $\sqrt{119}$ cm.

Q4. Draw a circle, then a tangent and a secant both parallel to a given line — a construction. Draw the given line; the tangent is the parallel line just touching the circle, the secant is a parallel line cutting it in two points.

9. NCERT Exercise 10.2 — fully solved (Q1–Q7)

Q1. Tangent length from $Q$ is $24$ cm and $OQ = 25$ cm. Radius $= \sqrt{OQ^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7$ cm. Answer: (A) $7$ cm.

Q2. $TP, TQ$ are tangents with $\angle POQ = 110^\circ$. In quadrilateral $OPTQ$, $\angle OPT = \angle OQT = 90^\circ$, and angles sum to $360^\circ$, so $\angle PTQ = 360^\circ - 90^\circ - 90^\circ - 110^\circ = 70^\circ$. Answer: (B) $70^\circ$.

Q3. Tangents $PA, PB$ inclined at $80^\circ$, i.e. $\angle APB = 80^\circ$. In quadrilateral $OAPB$, $\angle OAP = \angle OBP = 90^\circ$, so $\angle AOB = 360^\circ - 90^\circ - 90^\circ - 80^\circ = 100^\circ$. In right $\triangle OAP$, $\angle POA = 90^\circ - \dfrac{1}{2}\angle APB = 90^\circ - 40^\circ = 50^\circ$. Answer: (A) $50^\circ$.

Q4. Prove tangents at the ends of a diameter are parallel. Let $AB$ be a diameter; let the tangents at $A$ and $B$ be lines $\ell_A$ and $\ell_B$. By Theorem 10.1, $OA \perp \ell_A$ and $OB \perp \ell_B$. Since $A, O, B$ are collinear (diameter), $\ell_A$ and $\ell_B$ are both perpendicular to the same line $AB$, hence parallel. $\blacksquare$

Q5. Prove the perpendicular at the point of contact passes through the centre. Let tangent touch the circle at $P$, with centre $O$. By Theorem 10.1, $OP \perp$ tangent. So the perpendicular to the tangent at $P$ is the line $OP$ itself, which passes through the centre $O$. $\blacksquare$

Q6. Tangent length $4$ cm from point $A$ at distance $5$ cm from centre. Radius $= \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$ cm.

Q7. Concentric circles of radii $5$ cm and $3$ cm. The chord of the larger circle touching the smaller is bisected at the point of contact $P$ (Example 1), and $OP = 3$ is $\perp$ to it. Half-chord $= \sqrt{5^2 - 3^2} = \sqrt{16} = 4$ cm, so the full chord $= 2 \times 4 = 8$ cm.

10. NCERT Exercise 10.2 — fully solved (Q8–Q13)

Q8. Quadrilateral $ABCD$ circumscribes a circle. Prove $AB + CD = AD + BC$. Let the circle touch $AB, BC, CD, DA$ at $P, Q, R, S$. Equal tangents from each vertex (Theorem 10.2): $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$. Adding:

$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$, i.e. $AB + CD = AD + BC$. $\blacksquare$

Q9. $XY$ and $X'Y'$ are parallel tangents; another tangent $AB$ touches at $C$, cutting $XY$ at $A$, $X'Y'$ at $B$. Prove $\angle AOB = 90^\circ$. $A$ is external with tangents $AP$ (on $XY$) and $AC$, so $OA$ bisects $\angle PAC$; similarly $OB$ bisects $\angle QBC$. Since $XY \parallel X'Y'$, $\angle PAC + \angle QBC = 180^\circ$ (co-interior). Halving, $\angle OAC + \angle OBC = 90^\circ$. In $\triangle AOB$, $\angle AOB = 180^\circ - (\angle OAC + \angle OBC) = 180^\circ - 90^\circ = 90^\circ$. $\blacksquare$

Q10. Prove the angle between two tangents from an external point is supplementary to the angle subtended by the chord of contact at the centre. For external point $P$ with tangents $PA, PB$, quadrilateral $OAPB$ has $\angle OAP = \angle OBP = 90^\circ$. So $\angle APB + \angle AOB = 360^\circ - 90^\circ - 90^\circ = 180^\circ$ — the two angles are supplementary. $\blacksquare$

Q11. Prove a parallelogram circumscribing a circle is a rhombus. Let $ABCD$ be the parallelogram. By Q8, $AB + CD = AD + BC$. In a parallelogram $AB = CD$ and $AD = BC$, so $2AB = 2BC$, giving $AB = BC$. Adjacent sides equal $\Rightarrow$ all four sides equal $\Rightarrow$ rhombus. $\blacksquare$

Q12. Triangle $ABC$ circumscribes a circle of radius $4$ cm; the point of contact $D$ on $BC$ gives $BD = 8$ cm, $DC = 6$ cm. By equal tangents, let the tangent from $A$ be $x$: then $AE = AF = x$, $BD = BF = 8$, $CD = CE = 6$. So $AB = x + 8$, $AC = x + 6$, $BC = 14$. Using area two ways with incentre $O$ (radius $r = 4$):

Sides $a = BC = 14$, $b = CA = x + 6$, $c = AB = x + 8$; semiperimeter $s = \dfrac{14 + (x+6) + (x+8)}{2} = x + 14$.

Area by Heron $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)}$. Also Area $= r \cdot s = 4(x + 14)$.

So $48x(x+14) = 16(x+14)^2 \Rightarrow 48x = 16(x+14) \Rightarrow 3x = x + 14 \Rightarrow x = 7$.

Hence $AB = 7 + 8 = 15$ cm and $AC = 7 + 6 = 13$ cm.

Q13. Prove opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre. Let $ABCD$ touch the circle (centre $O$) at $P, Q, R, S$ on $AB, BC, CD, DA$. Join $O$ to the four vertices and four points of contact. The two tangents from each vertex subtend equal angles at $O$: write the eight small angles around $O$ as $\angle1,\dots,\angle8$ with $\angle1 = \angle2$, $\angle3 = \angle4$, $\angle5 = \angle6$, $\angle7 = \angle8$, and $\angle1+\dots+\angle8 = 360^\circ$. Grouping the angles subtended by opposite sides gives $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$. Hence opposite sides subtend supplementary angles at the centre. $\blacksquare$

11. Common mistakes to avoid

  • Forgetting that the radius is perpendicular to the tangent only at the point of contact — not at any other point.
  • Saying a circle has "two tangents" — it has two tangents from an external point, but infinitely many overall.
  • Using $PQ = \sqrt{OQ^2 - OP^2}$ with $OP$ and $OQ$ swapped — the hypotenuse is the line to the centre ($OQ$), and $OP$ = radius is a leg.
  • In quadrilateral $OAPB$ angle questions, forgetting the angle sum is $360^\circ$, not $180^\circ$.
  • Writing the tangent-length formula as $\sqrt{OP^2 + r^2}$ instead of $\sqrt{OP^2 - r^2}$.

12. Quick revision checklist

  • Line vs circle: 0 points (non-intersecting), 1 (tangent), 2 (secant).
  • Theorem 10.1: radius $\perp$ tangent at the point of contact.
  • Theorem 10.2: tangents from an external point are equal; centre bisects the angle between them.
  • Tangent length $= \sqrt{d^2 - r^2}$, where $d$ = distance from external point to centre.
  • From outside: 2 tangents; on the circle: 1; inside: 0.
  • $\angle$ between tangents $+\ \angle$ at centre (chord of contact) $= 180^\circ$.
Practice MCQs
1. A line that touches a circle at exactly one point is called a:
  1. secant
  2. chord
  3. tangent
  4. diameter
Answer: (C) tangent — it meets the circle in exactly one point, the point of contact.
2. The angle between a tangent and the radius at the point of contact is:
  1. $45^\circ$
  2. $60^\circ$
  3. $90^\circ$
  4. $180^\circ$
Answer: (C) $90^\circ$ — by Theorem 10.1 the radius is perpendicular to the tangent.
3. The length of a tangent from a point $25$ cm from the centre of a circle of radius $7$ cm is:
  1. $24$ cm
  2. $18$ cm
  3. $\sqrt{674}$ cm
  4. $32$ cm
Answer: (A) $\sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24$ cm.
4. The number of tangents that can be drawn from a point lying inside a circle is:
  1. $0$
  2. $1$
  3. $2$
  4. infinitely many
Answer: (A) $0$ — every line through an interior point cuts the circle in two points.
5. From an external point, the number of tangents to a circle is:
  1. $0$
  2. $1$
  3. $2$
  4. $3$
Answer: (C) $2$ — exactly two tangents can be drawn from a point outside the circle.
6. If two tangents from external point $P$ make an angle of $60^\circ$, the angle subtended by the chord of contact at the centre is:
  1. $60^\circ$
  2. $90^\circ$
  3. $120^\circ$
  4. $150^\circ$
Answer: (C) $120^\circ$ — the two angles are supplementary: $180^\circ - 60^\circ = 120^\circ$.
7. $PA$ and $PB$ are tangents from $P$ with $PA = 7$ cm. Then $PB$ equals:
  1. $3.5$ cm
  2. $7$ cm
  3. $14$ cm
  4. cannot be found
Answer: (B) $7$ cm — tangents from an external point are equal (Theorem 10.2).
8. The maximum number of parallel tangents a circle can have is:
  1. $1$
  2. $2$
  3. $3$
  4. infinitely many
Answer: (B) $2$ — one on each side, e.g. the tangents at the two ends of a diameter.
9. A quadrilateral $ABCD$ circumscribes a circle. If $AB = 6$, $BC = 7$, $CD = 4$, then $AD$ equals:
  1. $3$
  2. $5$
  3. $9$
  4. $11$
Answer: (A) $3$ — since $AB + CD = AD + BC$, $6 + 4 = AD + 7 \Rightarrow AD = 3$.
10. The tangents drawn at the ends of a diameter of a circle are:
  1. perpendicular
  2. intersecting
  3. parallel
  4. equal in length
Answer: (C) parallel — both are perpendicular to the same diameter, hence parallel.
Assertion–Reason
A: The tangents drawn from an external point to a circle are equal in length.   R: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Answer: Both A and R are true, and R is the correct explanation of A — Theorem 10.1 provides the right angles ($90^\circ$) that make the RHS congruence in Theorem 10.2 work, giving equal tangents.
A: A tangent to a circle meets it at exactly one point.   R: A secant to a circle meets it at exactly one point.
Answer: A is true, R is false — a secant meets the circle at two points; only a tangent meets it at one.
Previous-year questions
Q1. Prove that the lengths of tangents drawn from an external point to a circle are equal. (CBSE, 3 marks)
Outline: tangents $PQ, PR$ from $P$; join $OP, OQ, OR$. Right angles at $Q, R$ (Theorem 10.1); $OQ = OR$ (radii), $OP$ common $\Rightarrow \triangle OQP \cong \triangle ORP$ (RHS) $\Rightarrow PQ = PR$ (CPCT).
Q2. Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$ with $\angle POQ = 110^\circ$. Find $\angle PTQ$. (CBSE, 2 marks)
Answer: In quadrilateral $OPTQ$, $\angle OPT = \angle OQT = 90^\circ$, so $\angle PTQ = 360^\circ - 90^\circ - 90^\circ - 110^\circ = 70^\circ$.
Q3. A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$. (CBSE, 3 marks)
Outline: with contact points $P,Q,R,S$, use equal tangents from each vertex ($AP=AS$, $BP=BQ$, $CR=CQ$, $DR=DS$) and add to get $AB + CD = AD + BC$.
Q4. The length of a tangent from a point $A$ at distance $5$ cm from the centre is $4$ cm. Find the radius of the circle. (CBSE, 2 marks)
Answer: radius $= \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$ cm (right angle at the point of contact).
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