K2KARAKORAM
मराठीEnglish
Open the app

State Board · Class 10 · Mathematics Part 2

Lesson 3: Circle

5 concepts · 40 practice questions with answers · Free to read

This lesson is also in: मराठी

Circles through points, secant and tangent

From daily life

A cyclist rides along a straight road next to a round pond. The road touches the pond at exactly one spot and never enters it. A boat crossing the pond in a straight line enters at one spot and leaves at another.

Simple definition

A line meeting a circle in two points is a secant (छेदिका). A line meeting it in exactly one point is a tangent (स्पर्शिका), and that point is the point of contact (स्पर्शबिंदू).

Rule 1: Circles through points — Through one point, infinitely many circles pass. Through two points A and B, infinitely many pass, with centres on the perpendicular bisector of AB. Through three non-collinear points, exactly one circle passes. Three collinear points lie on no circle.

A circle with a secant cutting it at two points and a tangent touching at A, perpendicular to the radius OAOAsecanttangentradiusThe tangent touches at one point and is ⊥ to the radius

In the figure : In the figure a secant meets a circle in two points; a tangent in one point, and it is perpendicular to the radius there.

  • Easy : Easy: One circle passes through three corners of a triangle (its circumcircle).
  • Medium : Medium: Points A, B, C on a line: no circle passes through all three.
  • Exam-type : Exam: Why do centres of circles through A and B lie on one line? Every centre is equally far from A and B.

Rule 2: Tangent theorems — A tangent is perpendicular to the radius at the point of contact. Conversely, a line perpendicular to a radius at its end on the circle is a tangent. Tangent segments from an outside point are equal. Through an outside point, two tangents; through a point on the circle, one; through an inside point, none.

Two tangent segments PA and PB from an outside point P to a circle are equalOABPPA = PB = √(OP² − r²)

In the figure : In the figure the tangent segments from an outside point are equal.

  • Easy : Easy: Radius 5, centre to external point 13: tangent length = √(169 − 25) = 12.
  • Medium : Medium: Tangents from C make ∠ACB = 70°, so ∠ADB = 360° − (70° + 90° + 90°) = 110°.
  • Exam-type : Exam: Tangent length 15 from a point 17 cm from the centre gives radius √(289 − 225) = 8 cm. Two parallel tangents of a circle of radius 7 are 14 apart.
Remember

Tangent ⊥ radius. Equal tangent segments from an outside point.

Common mistake

Forgetting the right angle at the point of contact. Use it to form a right triangle.

Shortcut

Memory clue: "स्पर्श म्हणजे काटकोन".

Spot the difference

LineCommon points
Secant2
Tangent1
Outside line0

Practice questions

Easy
  1. 1Define a tangent.
  2. 2How many circles pass through three non-collinear points?
  3. 3Angle between tangent and radius at contact?
Medium
  1. 4Radius 3, tangent segment 4: distance from the centre?
  2. 5Distance between parallel tangents of a circle of radius 6.5?
  3. 6OR = 26 and radius 13: tangent segment and ∠ORM?
Challenging
  1. 7Tangents from P touch at A and B, ∠APB = 50°. Find ∠AOB.
  2. 8Prove that a circle touching all sides of a parallelogram makes it a rhombus.
Show answers (try them yourself first)
  1. 1A line meeting the circle in exactly one point.
  2. 2Exactly one.
  3. 390°.
  4. 45.
  5. 513.
  6. 613√3; 13 is half of 26, so ∠ORM = 30°.
  7. 7360° − (50° + 90° + 90°) = 130°.
  8. 8Tangent segments from each vertex are equal, so AB + CD = AD + BC, which with opposite sides equal gives AB = AD.

Touching circles

From daily life

Two coins pressed together on a table touch at one point. A small coin inside a large ring touching its edge also meets it at one point. These are touching circles.

Simple definition

Two circles touching each other have a common tangent at the point of contact. They touch externally (बाह्य) or internally (अंतर्गत).

Rule 1: Point of contact and distance — The point of contact lies on the line joining the centres. Externally touching circles: distance between centres = r1 + r2. Internally touching: distance = r1 − r2. Radii 7 and 2: 9 (externally) or 5 (internally).

Two circles touching externally with centres 5 apart and two circles touching internally with centres 3 apartThe point of contact lies on the line of centresABC32Outside contact:AB = 3 + 2 = 5MNTInside contact:MN = 5 − 2 = 3Centres: sum of radii (outside), difference (inside)

In the figure : In the figure the distance between the centres of touching circles: the sum of the radii (outside) or the difference (inside).

  • Easy : Easy: Radii 6 and 15 touching internally: centres 9 apart.
  • Medium : Medium: Three circles of radius 4 touching each other have centres forming an equilateral triangle of side 8, perimeter 24.
  • Exam-type : Exam: Externally touching circles with one radius 5 and centres 17 apart: the other radius is 12.

Rule 2: Common tangent length — Radii 4 and 9 touching externally: d = 13. A direct common tangent has length √(d2 − (9 − 4)2) = √(169 − 25) = 12. Two externally touching circles have 3 common tangents.

  • Easy : Easy: Radii 3 and 3 touching externally: d = 6.
  • Medium : Medium: Radii 2 and 8 touching externally: d = 10, tangent = √(100 − 36) = 8.
  • Exam-type : Exam: If two circles each pass through the other's centre with centres 12 apart, each radius is 12.
Remember

Outside: add the radii. Inside: subtract. The contact point lies on the line of centres.

Common mistake

Adding radii for internal contact. For internal contact, subtract.

Shortcut

Memory clue: "बाह्य = बेरीज, अंतर्गत = फरक".

Spot the difference

TouchDistance between centres
Externalr1 + r2
Internalr1 − r2
Radii 7, 29 or 5

Practice questions

Easy
  1. 1What lies on the line joining the centres of touching circles?
  2. 2Distance between centres for external contact?
  3. 3Distance between centres for internal contact?
Medium
  1. 4Radii 5.5 and 3.3 touching externally: d?
  2. 5Radii 5.5 and 3.3 touching internally: d?
  3. 6How many common tangents do externally touching circles have?
Challenging
  1. 7Radii 4 and 9 touch externally. Find the length of the common tangent.
  2. 8Circles of radius 10 and 4 touch internally. Find d. If a point on the smaller one is on the line of centres, where is it?
Show answers (try them yourself first)
  1. 1The point of contact.
  2. 2Sum of the radii.
  3. 3Difference of the radii.
  4. 48.8.
  5. 52.2.
  6. 6Three.
  7. 712.
  8. 8d = 6. The contact point lies on the line of centres, beyond the smaller circle's centre.

Arc of a circle

From daily life

A pizza cut along a straight line gives a big piece and a small piece. The curved edges are arcs, and the angle at the middle is the central angle.

Simple definition

An arc (कंस) is a part of a circle. The central angle (केंद्रीय कोन) has its vertex at the centre. The smaller arc is the minor arc (लघुकंस), the larger is the major arc (विशालकंस).

Rule 1: Measure of an arc — Minor arc = measure of its central angle. Major arc = 360° − minor arc. Semicircle = 180°. Full circle = 360°. Central angle 80°: minor arc 80°, major arc 280°.

A circle with a minor arc AB of 70 degrees and the major arc of 290 degreesOAB70°minor arc AB = 70°major arc = 360° − 70°= 290°Minor arc = itscentral angle

In the figure : In the figure a minor arc has the measure of its central angle; the major arc is 360° minus that.

  • Easy : Easy: Central angle 120°: minor arc 120°, major arc 240°.
  • Medium : Medium: m(arc AB) = 105°, m(arc BC) = 140°: m(arc ABC) = 245°, m(arc AC) = 360° − 245° = 115°.
  • Exam-type : Exam: A chord equal to the radius makes an equilateral triangle with the centre, so its minor arc is 60° and major arc 300°.

Rule 2: Congruent arcs — Equal measure is not enough; congruent arcs need equal measure and equal radii. Congruent arcs have congruent chords, and congruent chords have congruent arcs. Arcs add only if they share just one point: m(arc ABC) + m(arc CDE) = m(arc ACE).

  • Easy : Easy: An equilateral triangle in a circle makes three arcs of 120° each.
  • Medium : Medium: Chord AB ≅ chord CD, so arc AB ≅ arc CD.
  • Exam-type : Exam: In a rectangle inscribed in a circle, opposite sides are congruent chords, so arc PQ ≅ arc SR.
Remember

Minor arc = central angle. Major arc = 360° − minor arc.

Common mistake

Calling arcs congruent just because measures are equal. Radii must be equal too.

Shortcut

Memory clue: "लघुकंस = कोन; विशाल = 360 − कोन".

Spot the difference

ArcMeasure
Minor= central angle
Major360° − minor
Semicircle180°

Practice questions

Easy
  1. 1What is a central angle?
  2. 2Measure of a semicircle?
  3. 3Central angle 65°: major arc?
Medium
  1. 4m(arc AB) = 95°, m(arc BC) = 130°, no other overlap. Find m(arc ABC).
  2. 5Equal chords make what kind of arcs?
  3. 6m(arc BAC) if m(arc BC) = 140°?
Challenging
  1. 7Equilateral ΔQRS inscribed in a circle: m(arc QRS)?
  2. 8Why can we not add m(arc ABC) and m(arc BCE) to get m(arc ABE)?
Show answers (try them yourself first)
  1. 1An angle with its vertex at the centre.
  2. 2180°.
  3. 3295°.
  4. 4225°.
  5. 5Congruent arcs.
  6. 6220°.
  7. 7240°.
  8. 8They overlap in many points.

Inscribed angle and cyclic quadrilateral

From daily life

Standing at different spots on the edge of a round stage, you still see the same two lamps at the same angle, as long as you stay on the same side. That is the inscribed angle idea.

Simple definition

An inscribed angle (अंतर्लिखित कोन) has its vertex on the circle. It equals half the arc it intercepts. A cyclic quadrilateral (चक्रीय चौकोन) has all four vertices on a circle.

Rule 1: Inscribed angle theorem — Inscribed angle = 12 × intercepted arc = 12 × central angle on that arc. Arc 84°: inscribed angle 42°. Inscribed angle 36°: arc 72°. Same arc: equal angles. Angle in a semicircle = 90°.

The inscribed angle ACB is 50 degrees, half of the 100 degree arc AB and half of the central angle100°50°OABCarc AB = 100°central ∠AOB= 100°inscribed ∠ACB= ½ × 100° = 50°

In the figure : In the figure the inscribed angle is half of the arc it intercepts.

  • Easy : Easy: Central angle 100° gives inscribed angle 50°.
  • Medium : Medium: In ΔABC with AB a diameter and ∠A = 35°: ∠C = 90°, ∠B = 55°.
  • Exam-type : Exam: ∠ACB = 65° inscribed in arc ACB: the intercepted arc is 130°, so m(arc ACB) = 360° − 130° = 230°.

Rule 2: Cyclic quadrilateral — Opposite angles of a cyclic quadrilateral are supplementary: ∠A + ∠C = 180° and ∠B + ∠D = 180°. If ∠A = 75°, then ∠C = 105°. The exterior angle equals the opposite interior angle. Converse: if opposite angles add to 180°, the quadrilateral is cyclic. Every rectangle is cyclic.

A cyclic quadrilateral ABCD with opposite angles adding to 180 degrees∠A∠C∠B∠DABCD∠A + ∠C = 180°∠B + ∠D = 180°Opposite angles of acyclic quadrilateralare supplementary

In the figure : In the figure in a cyclic quadrilateral the opposite angles add up to 180°.

  • Easy : Easy: ∠B = 110° gives ∠D = 70°.
  • Medium : Medium: ∠A = (3x + 10)°, ∠C = (2x + 20)°: 5x + 30 = 180, x = 30, so ∠A = 100° and ∠C = 80°.
  • Exam-type : Exam: Interior ∠C = 105° means the exterior angle at C is 75° = ∠A.

Rule 3: Angle between chords and secants — Chords meeting inside: angle = 12 (sum of the two intercepted arcs). With arcs 40° and 80°: 60°. Secants meeting outside: angle = 12 (difference of arcs). With arcs 100° and 30°: 35°.

Live figure

Inscribed angle and cyclic quadrilateral

In the app you can move this figure with your finger.

Open in the app
  • Easy : Easy: Arcs 50° and 70° inside give 12 × 120° = 60°.
  • Medium : Medium: Arcs 120° and 40° outside give 12 × 80° = 40°.
  • Exam-type : Exam: Chords WX and YZ meet at T, arcs 44° and 68°: ∠ZTX = 12 × 112° = 56°.
Remember

Inscribed angle = half the arc. Cyclic quadrilateral: opposite angles add to 180°.

Common mistake

Using the minor arc when the angle faces the major arc. Check which arc is intercepted.

Shortcut

Memory clue: "कोन = कंस ÷ 2; चक्रीय → 180°".

Spot the difference

RuleResult
Inscribed angle12 arc
Angle in semicircle90°
Cyclic quadrilateralOpposite angles = 180°

Practice questions

Easy
  1. 1State the inscribed angle theorem.
  2. 2Angle in a semicircle?
  3. 3Arc 110°: inscribed angle?
Medium
  1. 4∠A of a cyclic quadrilateral is 80°. Find ∠C.
  2. 5Is a rectangle cyclic? Why?
  3. 6∠R = (5x − 13)° and ∠N = (4x + 4)° of cyclic MRPN. Find both.
Challenging
  1. 7Chord equals the radius. Find the inscribed angle on the major arc.
  2. 8In a cyclic quadrilateral 2∠A = 3∠C. Find ∠C.
Show answers (try them yourself first)
  1. 1The angle is half the intercepted arc.
  2. 290°.
  3. 355°.
  4. 4100°.
  5. 5Yes, opposite angles are 90° + 90° = 180°.
  6. 69x − 9 = 180, x = 21, ∠R = 92°, ∠N = 88°.
  7. 7Central angle 60°, inscribed angle 30°.
  8. 8∠A = 1.5∠C, 2.5∠C = 180°, ∠C = 72°.

Secant tangent angle theorem and theorem of intersecting chords

From daily life

A cricket ball is thrown from a point outside a round pond and hits the water twice, once on entering and once on leaving. The distance products from the outside point to the two splash points follow a precise rule.

Simple definition

Chords AB and CD meeting at E satisfy AE × EB = CE × ED. For a tangent ET and a secant EAB from E: ET2 = EA × EB.

Rule 1: Tangent–chord angle and internal chords — The angle between a tangent and a chord = 12 × the intercepted arc. Arc 110°: angle 55°. Angle 40°: arc 80°. If chords AB and CD meet inside at E: AE × EB = CE × ED. With AE = 6, EB = 4, CE = 3: ED = 243 = 8.

Two chords AB and CD of a circle crossing at E: AE times EB equals CE times ED, both 24EABCDAE = 6, EB = 4CE = 3, ED = 8AE × EB = 24CE × ED = 24The two productsare equal

In the figure : In the figure intersecting chords: AE × EB = CE × ED (new numbers, drawn to scale).

  • Easy : Easy: Chords meet at E with parts 2 and 12, and 4 and x: x = 6.
  • Medium : Medium: Chord PQ = 10 split at E so that PE × EQ = 16: x(10 − x) = 16, x = 2 or 8.
  • Exam-type : Exam: Why does AE × EB = CE × ED? ΔCAE ~ ΔBDE by AA, so AEDE = CEBE.

Rule 2: Tangent–secant and secant–secant — A tangent PT = 10 and a secant PQR with PQ = 4: PR = 1004 = 25 and QR = 21. Two secants from E: EA × EB = EC × ED with EA = 9, EB = 4, EC = 12, ED = 3, both 36. Tangent 15 and secant outer part 9: whole secant = 2259 = 25, chord = 16.

A tangent PT of 4 and a secant PQR with PQ 2 and PR 8: PT squared equals PQ times PROPTQRPT = 4PQ = 2, PR = 8PT² = PQ × PR16 = 2 × 8 ✓

In the figure : In the figure tangent–secant: PT2 = PQ × PR.

  • Easy : Easy: Tangent 6, outer part 3: whole secant = 363 = 12, chord 9.
  • Medium : Medium: Tangent 12, whole secant 18: outer part = 14418 = 8, chord 10.
  • Exam-type : Exam: Two circles meet at X and Y; tangents from M on line XY touch them at P and Q. PM2 = MX × MY = QM2, so PM = QM.
Remember

Chords inside: part × part. Tangent² = outer × whole.

Common mistake

Using only the chord length instead of the whole secant from the outside point.

Shortcut

Memory clue: "भाग×भाग = भाग×भाग; स्पर्श² = बाह्य×पूर्ण".

Spot the difference

SituationRule
Chords insideAE × EB = CE × ED
Two secants outsideEA × EB = EC × ED
Tangent + secantET2 = EA × EB

Practice questions

Easy
  1. 1State the intersecting chords rule.
  2. 2State the tangent–secant rule.
  3. 3AE = 5, EB = 4, CE = 2. Find ED.
Medium
  1. 4Tangent 8, outer part 4: whole secant?
  2. 5Tangent–chord angle for arc 130°?
  3. 6PS ⊥ diameter PQ with R on the circle (S on PQ). PS = 4, SQ = 9. Find SR.
Challenging
  1. 7Tangent PQ = 12, secant PRS with PR = 8. Find PS and RS.
  2. 8Secants from A meet arcs 100° and 40°. Find the angle at A.
Show answers (try them yourself first)
  1. 1AE × EB = CE × ED.
  2. 2ET² = EA × EB.
  3. 310.
  4. 416.
  5. 565°.
  6. 6SR² = 36, SR = 6.
  7. 7PS = 18, RS = 10.
  8. 812 (100° − 40°) = 30°.

Quick revision

  • The tangent at a point of a circle is perpendicular to the radius at that point (and a line perpendicular to a radius at its point on the circle is a tangent); tangent segments from an external point are congruent; a unique circle passes through three non-collinear points; for touching circles the distance between the centres is the sum of the radii (externally) or the difference of the radii (internally).
  • The measure of a minor arc is the measure of its central angle and that of a major arc is 360° minus it; an inscribed angle is half of its intercepted arc, angles in the same arc are congruent and an angle in a semicircle is 90°; opposite angles of a cyclic quadrilateral are supplementary.
  • The angle between a tangent and a chord is half of the intercepted arc; for chords or secants meeting at E, AE × EB = CE × ED; for a tangent segment ET and a secant EAB from an external point, ET² = EA × EB.