Chapter 6 — Circle (Class 9) — Quick Q&A & Textbook Solutions

20 Most Important — 1 Mark Questions (Answers in green)

Q1. Define a circle.
A1. A circle is the set of all points in a plane equidistant from a fixed point called the centre.
Q2. What is the radius?
A2. Radius is the segment joining the centre of the circle to any point on the circle.
Q3. What is a chord?
A3. A chord is a segment joining any two points on the circle.
Q4. What is a diameter?
A4. A diameter is a chord passing through the centre. It is the longest chord; Diameter = 2 × radius.
Q5. State the theorem about perpendicular from centre to chord.
A5. A perpendicular drawn from the centre to a chord bisects the chord.
Q6. If two chords are congruent, what can you say about their distances from the centre?
A6. Congruent chords are equidistant from the centre.
Q7. Converse: if two chords are equidistant from the centre what follows?
A7. Chords equidistant from the centre are congruent (have equal lengths).
Q8. Where does the circumcentre lie for a right triangle?
A8. The circumcentre is the midpoint of the hypotenuse (lies on the hypotenuse).
Q9. Where does the incentre lie for any triangle?
A9. The incentre always lies inside the triangle.
Q10. What is the incircle of a triangle?
A10. Incircle is the circle that touches all three sides of the triangle internally; its centre is the intersection of angle bisectors.
Q11. What are concentric circles?
A11. Concentric circles have the same centre but different radii.
Q12. What is the relationship between radius and diameter numerically?
A12. Diameter \(=2r\).
Q13. If a chord equals the diameter, what type of chord is it?
A13. It is the largest chord — the diameter itself.
Q14. If a perpendicular from centre meets chord at its midpoint, what can be inferred?
A14. The perpendicular from the centre passes through the midpoint of the chord (bisects it).
Q15. In an isosceles triangle, where lies the circumcentre relative to axis of symmetry?
A15. The circumcentre lies on the axis of symmetry (perpendicular bisector of base).
Q16. For an equilateral triangle, what is special about incentre and circumcentre?
A16. They coincide (same point); perpendicular bisectors and angle bisectors all coincide.
Q17. If distance from centre to chord is equal to radius, what is chord length?
A17. If distance \(=r\) then half-chord length is \(\sqrt{r^2-r^2}=0\). So chord length \(=0\) (i.e., point). Practically this means chord degenerates to a point (on the circle).
Q18. What is the length of the longest chord of a circle of radius \(2.9\) cm?
A18. Longest chord = diameter \(=2 \times 2.9 = 5.8\) cm.
Q19. What is the locus definition of circle in one sentence?
A19. It is the locus of points at fixed distance (radius) from a fixed point (centre).
Q20. If a line through the centre meets the circle at two points, what is that line called?
A20. That line segment inside the circle joining the two intersection points through the centre is a diameter.

20 Most Important — 2 Marks Questions

Q1. State and prove: A perpendicular from the centre to a chord bisects the chord.
A1. Proof (brief): Let circle centre \(O\). Let \(AB\) be chord and \(OP\perp AB\) meet at \(P\). Join \(OA\) and \(OB\). Triangles \(OPA\) and \(OPB\) are right triangles with \(OP\) common, \(OA=OB\) (radii). So by RHS, \(\triangle OPA\cong\triangle OPB\). Hence \(PA=PB\). So \(OP\) bisects \(AB\).
Q2. Show: The segment joining centre and midpoint of chord is perpendicular to the chord.
A2. If \(P\) is midpoint of chord \(AB\), then \(AP=PB\). Join \(OA\) and \(OB\). Triangles \(OAP\) and \(OBP\) have \(OA=OB\), \(AP=PB\), \(OP\) common — so by SSS they are congruent. Hence \(\angle OPA=\angle OPB\). They sum to \(180^\circ\) (linear pair), so each is \(90^\circ\). Thus \(OP\perp AB\).
Q3. If radius \(r=5\) cm and chord length \(=8\) cm, find distance of chord from centre.
A3. Half chord \(=4\). In right triangle with hypotenuse \(=5\) and half-chord \(=4\): distance \(d=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3\) cm.
Q4. Radius \(=20\) cm, distance from centre to chord \(=12\) cm. Find chord length.
A4. Half-chord \(= \sqrt{20^2-12^2}=\sqrt{400-144}=\sqrt{256}=16\). So chord length \(=32\) cm.
Q5. Show that congruent chords are equidistant from centre.
A5. Let \(AB=CD\), let perpendiculars from \(O\) meet them at \(P,Q\). Then \(AP= \tfrac{1}{2}AB\) and \(DQ=\tfrac{1}{2}CD\) so \(AP=DQ\). In right triangles \(OPA\) and \(OQD\), hypotenuses \(OA=OD\), legs \(AP=DQ\) so by RHS congruence \(OP=OQ\). Thus distances equal.
Q6. Converse: If two chords are equidistant from the centre show they are congruent.
A6. Let distances \(OP=OQ\). In right triangles \(OPA\) and \(OQD\), \(OA=OD\) (radii), \(OP=OQ\) (given) so triangles congruent → \(AP=DQ\). Thus \(AB=CD\).
Q7. Radius \(=10\) cm, chord length \(=16\). Find distance from centre.
A7. Half-chord \(=8\). Distance \(= \sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6\) cm.
Q8. In triangle, point where angle bisectors meet is called?
A8. The incentre — point of concurrency of angle bisectors; it is equidistant from sides.
Q9. Explain briefly how to construct incircle of triangle \(ABC\).
A9. Construct bisectors of two angles, their intersection is incentre \(I\). Draw perpendicular from \(I\) to any side; that distance is radius. Draw circle with centre \(I\) and that radius.
Q10. Explain briefly how to construct circumcircle of triangle \(ABC\).
A10. Construct perpendicular bisectors of two sides; their intersection is circumcentre \(O\). Distance from \(O\) to any vertex is radius. Draw circle centre \(O\) radius \(OA\).
Q11. If chord length \(24\) and diameter \(26\), find distance from centre to chord.
A11. Radius \(=13\). Half-chord \(=12\). Distance \(= \sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\) cm.
Q12. If radius \(=34\) and distance to chord \(=30\), find chord length.
A12. Half-chord \(= \sqrt{34^2-30^2}=\sqrt{1156-900}=\sqrt{256}=16\). Chord length \(=32\).
Q13. If radius \(=41\) and chord \(=80\), find distance from centre.
A13. Half-chord \(=40\). Distance \(= \sqrt{41^2-40^2}=\sqrt{1681-1600}=\sqrt{81}=9\) units.
Q14. Show that if a diameter bisects two chords then those chords are parallel.
A14. Let diameter \(d\) meets chords \(AB\) and \(CD\) at their midpoints \(P,Q\). Then \(d\perp AB\) and \(d\perp CD\). Two lines perpendicular to the same line are parallel ⇒ \(AB\parallel CD\).
Q15. Two equal chords are drawn in a circle. Show their perpendicular distances from centre are equal (brief).
A15. Follows from the theorem: congruent chords are equidistant from the centre (proof in Q5).
Q16. If chord \(AB\) passes through point \(P\) on a smaller concentric circle and intersects bigger circle at points \(A,B\), show \(AP=BQ\) (figure context).
A16. In the figure where equal radii or symmetry exists, by drawing perpendiculars and congruent triangles one can show corresponding segments equal. (This is the textbook activity — using congruent triangles one proves \(AP=BQ\).)
Q17. If a chord is at distance 4 cm from centre and another congruent chord is at distance 4 cm, what is the length relation?
A17. Their lengths are equal (congruent) by converse theorem.
Q18. Radius \(=13\), two equal chords at distance \(5\) from centre. Find chord lengths.
A18. Half-chord \(= \sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\). Chord length \(=24\).
Q19. In right triangle, where is circumcentre located?
A19. At midpoint of hypotenuse — equidistant from all three vertices.
Q20. In an equilateral triangle of side \(a\) the ratio of circumradius to inradius is?
A20. Ratio \(R:r = 2:1\).

20 Most Important — 3 Marks Questions

Q1. Prove: A perpendicular from the centre to a chord bisects the chord, complete proof with congruence.
A1. Let \(O\) be centre, chord \(AB\), \(OP\perp AB\) at \(P\). Join \(OA,OB\). In \(\triangle OPA\) and \(\triangle OPB\): \(OA=OB\) (radii), \(OP\) common, \(\angle OPA=\angle OPB=90^\circ\). So triangles congruent by RHS ⇒ \(PA=PB\). Hence \(OP\) bisects \(AB\).
Q2. A chord of length \(12\) cm is at distance \(8\) cm from centre. Find diameter.
A2. Half-chord \(=6\). Using \(r^2 = d^2 + \text{(half-chord)}^2\) where \(d=8\) is distance. So \(r=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\). Diameter \(=20\) cm.
Q3. In the figure two congruent chords \(PM,PN\) of circle centre \(C\). Show ray \(PC\) bisects angle \(NPM\).
A3. Since chords \(PM\) and \(PN\) congruent, their perpendicular distances from \(C\) are equal; joining \(PC\) to midpoints gives right triangles symmetric about \(PC\). So \(PC\) bisects the angle between the chords (equal corresponding angles in congruent triangles).
Q4. Given diameter \(=26\), chord \(=24\). Find distance of chord from centre (show working).
A4. Radius \(=13\). Half-chord \(=12\). Distance \(= \sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\) cm.
Q5. Prove: The chords equidistant from centre are congruent (complete steps).
A5. Let \(AB\) and \(CD\) be chords with \(OP=OQ\) where \(P,Q\) are feet of perpendiculars. Join \(OA,OD\). In right triangles \(OPA\) and \(OQD\): \(OA=OD\) (radii), \(OP=OQ\) (given) → triangles congruent by RHS ⇒ \(AP=DQ\) so \(AB=CD\).
Q6. Radius of circle \(=10\) cm; two chords each length \(16\) cm. Find distance of these chords from centre.
A6. Half-chord \(=8\). Distance \(= \sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6\) cm.
Q7. Find the length of a chord at distance \(3\) cm from centre of circle radius \(5\) cm.
A7. Half-chord \(= \sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4\). Chord length \(=8\) cm.
Q8. In triangle \(ABC\), angle bisectors meet at \(I\). Explain why \(I\) is equidistant from sides.
A8. Any point on angle bisector is equidistant from the sides of that angle. As \(I\) lies on all three bisectors, it is equidistant from all three sides: \(IP=IQ=IR\).
Q9. Construct incircle of triangle with given sides/angles — outline steps and why it works.
A9. Bisect two angles; intersection \(I\) is incentre. Perpendicular from \(I\) to a side gives radius. Circle centre \(I\), that radius touches all three sides because each side is at same perpendicular distance from \(I\).
Q10. Construct circumcircle of triangle — outline steps and reason.
A10. Perpendicular bisectors of two sides meet at circumcentre \(O\). Distance \(OA\) is radius; circle through vertices passes all three because perpendicular bisector points are equidistant from endpoints.
Q11. Radius of circle \(=13\), two equal chords are at distance \(5\) from centre. Find chord lengths (show math).
A11. Half-chord \(= \sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\) → chord \(=24\).
Q12. Explain why the perpendicular bisectors and angle bisectors coincide for an equilateral triangle.
A12. In equilateral triangle all sides equal, so perpendicular bisector of any side also bisects opposite angle (symmetry) — hence they coincide.
Q13. In a circle centre \(O\), chord \(AB\) at distance \(d\) from \(O\) has length \(l\). Derive relation between \(r,d,l\).
A13. Half-chord \(= \tfrac{l}{2}\). By right triangle: \(r^2 = d^2 + \left(\tfrac{l}{2}\right)^2\). So \(l = 2\sqrt{r^2-d^2}\).
Q14. Show that the diameter is the longest chord of a circle.
A14. For any chord with half-length \(x\) at distance \(d\): \(r^2 = d^2 + x^2\) so \(x \le r\). Hence chord length \(=2x \le 2r\) and equality when \(d=0\) i.e. chord passes through centre (diameter).
Q15. A chord of length \(24\) is at distance \(5\) from centre. Find radius.
A15. Half-chord \(=12\). \(r=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\) cm.
Q16. If a diameter bisects a chord not through centre, show what happens.
A16. If diameter bisects a chord, it is perpendicular to the chord (because bisector from centre must be perpendicular), so diameter ⟂ chord.
Q17. Solve: In a circle radius \(=5.5\) cm, find a chord whose half-length equals \(4\) cm; find distance from centre.
A17. Distance \(= \sqrt{5.5^2-4^2}=\sqrt{30.25-16}=\sqrt{14.25}\approx 3.775\) cm.
Q18. Prove: If two equal chords are on opposite sides of the centre, the distance between them equals sum of their distances from centre.
A18. Let distances be \(d_1,d_2\); chords are on opposite sides so separation along the diameter through their perpendiculars equals \(d_1+d_2\) by straight-line geometry.
Q19. Given triangle is isosceles with equal sides as radii of circumcircle — what can be said about circumcentre?
A19. For isosceles triangle, circumcentre lies on axis of symmetry (perpendicular bisector of base).
Q20. A chord of length \(32\) in a circle of radius \(20\). Find its distance from centre (show steps).
A20. Half-chord \(=16\). Distance \(= \sqrt{20^2-16^2}=\sqrt{400-256}=\sqrt{144}=12\) cm.

Textbook Exercise Questions — Perfect Solutions (selected explicit questions from the provided text)

Practice Set 6.1 — Solutions

P6.1 — Q1. Distance of chord AB from centre = 8 cm; chord length = 12 cm. Find diameter.
Half-chord \(=6\). Radius \(r=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\) ⇒ diameter \(=20\) cm.
P6.1 — Q2. Diameter \(=26\) cm and chord length \(=24\) cm. Find distance of chord from centre.
Radius \(=13\). Half-chord \(=12\). Distance \(= \sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\) cm.
P6.1 — Q3. Radius \(=34\) and distance from centre \(=30\); find length of chord.
Half-chord \(= \sqrt{34^2-30^2}=\sqrt{1156-900}=\sqrt{256}=16\). Chord length \(=32\).
P6.1 — Q4. Radius \(=41\), chord PQ \(=80\). Find distance of chord from centre.
Half-chord \(=40\). Distance \(= \sqrt{41^2-40^2}=\sqrt{1681-1600}=\sqrt{81}=9\) units.
P6.1 — Q5 (figure 6.9). Centre \(O\) for two circles; chord \(AB\) of bigger circle intersects smaller circle at \(P,Q\). Show \(AP=BQ\).
Consider the two chords \(AP\) and \(BQ\) formed by the intersection. Using symmetry about the line through \(O\) (join O to midpoint), construct perpendiculars from \(O\) to the chord and observe congruent right triangles which give equal corresponding segments — so \(AP=BQ\). (Textbook figure uses congruence of triangles formed by radii and halves of chords.)
P6.1 — Q6. Prove that if a diameter bisects two chords of the circle then those two chords are parallel to each other.
Let diameter \(d\) intersect chords \(AB\) and \(CD\) at midpoints \(P\) and \(Q\). Then \(d\) is perpendicular to both \(AB\) and \(CD\) (perpendicular from centre bisects chord). Two lines perpendicular to same line are parallel. Therefore \(AB\parallel CD\).

Practice Set 6.2 — Solutions

P6.2 — Q1. Radius \(=10\) cm. Two chords of length \(16\) each. What is distance of these chords from centre?
Half-chord \(=8\). Distance \(= \sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6\) cm.
P6.2 — Q2. In circle radius \(=13\) cm, two equal chords are at distance \(5\) cm from centre. Find chord lengths.
Half-chord \(= \sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\). Chord \(=24\) cm.
P6.2 — Q3. Segments \(PM\) and \(PN\) are congruent chords of circle centre \(C\). Show ray \(PC\) is bisector of \(\angle NPM\).
Since chords are congruent their perpendicular distances from \(C\) are equal. Joining \(PC\) to midpoints produces two congruent right triangles symmetric about \(PC\). Hence angles at \(P\) adjacent to \(PC\) are equal ⇒ \(PC\) bisects \(\angle NPM\).

Practice Set 6.3 — Construction Steps (Concise)

P6.3 — Q1. Construct \(\triangle ABC\) with \(\angle B=100^\circ, BC=6.4\) cm, \(\angle C=50^\circ\) and construct its incircle.
Steps: (1) Draw base \(BC=6.4\) cm. (2) At \(B\) construct \(100^\circ\), at \(C\) construct \(50^\circ\). Their intersection gives \(A\). (3) Bisect \(\angle B\) and \(\angle C\); intersection \(I\) is incentre. (4) Drop perpendicular from \(I\) to \(BC\) meeting at \(M\). (5) With centre \(I\) and radius \(IM\), draw incircle.
P6.3 — Q2. Construct \(\triangle PQR\) with \(\angle P=70^\circ, \angle R=50^\circ, QR=7.3\) cm and its circumcircle.
Steps: (1) Draw base \(QR=7.3\). (2) At \(Q\) make \(180^\circ-70^\circ=110^\circ\) (internal) and at \(R\) make \(180^\circ-50^\circ=130^\circ\) to locate \(P\). (3) Draw perpendicular bisectors of \(QR\) and \(PR\); intersection \(O\) is circumcentre. (4) With centre \(O\) radius \(OQ\) draw circumcircle.
P6.3 — Q3. Construct \(\triangle XYZ\) with sides \(XY=6.7\), \(YZ=5.8\), \(XZ=6.9\); construct its incircle.
Steps: (1) Construct triangle using three sides. (2) Bisect any two angles and find incentre \(I\). (3) Drop perpendicular to a side for radius, draw circle centre \(I\) radius that length.
P6.3 — Q4. In \(\triangle LMN\), given \(LM=7.2\), \(\angle M=105^\circ\), \(MN=6.4\); construct circumcircle.
Steps: (1) Draw side \(MN=6.4\). (2) At \(M\) erect angle \(105^\circ\) and use side length \(LM=7.2\) to locate \(L\). (3) Perpendicular bisectors of two sides intersect at \(O\) — circumcentre. (4) With radius \(OL\) draw circle.
P6.3 — Q5. Construct \(\triangle DEF\) with \(DE=EF=6\), \(\angle F=45^\circ\) and its circumcircle.
Steps: (1) Construct isosceles triangle with base \(DF\) found via geometry using given conditions. (2) Perpendicular bisectors intersect at \(O\) — circumcentre. (3) Draw circumcircle radius \(OE\).

Problem Set 6 — Selected Solutions

Q1 (MCQ) — (ii) Radius \(=10\) cm, distance to chord \(=6\). Find chord length.
Half-chord \(= \sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\). So chord \(=16\) cm. Answer: (A) 16 cm.
Q1 (MCQ) — (ii extra) The point of concurrence of angle bisectors is called?
Incentre — (C).
Q1 (MCQ) — (iii) The circle passing through all vertices of triangle is called?
Circumcircle — (A).
Q1 (MCQ) — (iv) Length of chord \(=24\), distance from centre \(=5\). Find radius.
Half-chord \(=12\). Radius \(=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\). Answer: (B) 13 cm.
Q1 (MCQ) — (v) Longest chord for radius \(=2.9\) cm?
Longest chord = diameter \(=2\times2.9=5.8\) cm ⇒ (D) 5.8 cm.
Q1 (MCQ) — (vi) If \(OP=4.2\) and radius \(=4\) where is \(P\)?
Since \(OP>r\), point \(P\) is outside the circle. Answer: (C).
Q1 (MCQ) — (vii) Parallel chords on opposite sides have lengths \(6\) and \(8\), radius \(=5\). Find distance between them.
Let distances from centre be \(d_1,d_2\). For chord 6: half =3 ⇒ \(d_1^2 = 5^2-3^2=25-9=16\) ⇒ \(d_1=4\). For chord 8, half=4 ⇒ \(d_2^2=25-16=9\) ⇒ \(d_2=3\). On opposite sides distance between chords = \(d_1+d_2=4+3=7\) cm. Answer: (D) 7 cm.
Q2. Construct incircle and circumcircle of equilateral triangle side \(7.5\) cm. Measure radii and find ratio \(R:r\).
For equilateral triangle side \(a\): circumradius \(R=\dfrac{a}{\sqrt{3}}\), inradius \(r=\dfrac{a}{2\sqrt{3}}\). So \(R:r = 2:1\). (Concrete: \(R=\dfrac{7.5}{\sqrt3}\), \(r=\dfrac{7.5}{2\sqrt3}\).)
Q4 (figure 6.19). \(C\) centre, \(QT\) diameter, \(CT=13, CP=5\). Find length of chord \(RS\).
Interpretation & solution: We interpret \(CT\) as radius (so \(r=13\)). If \(P\) is a point at distance \(CP=5\) from centre, and \(RS\) is a chord at that distance (i.e., perpendicular from \(C\) to chord meets at distance 5), then half-chord \(= \sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\). So chord length \(=24\). (If the textbook figure intends something else, the same method—Pythagoras—applies.)
Q5 (figure 6.20). Chords \(AB\) and \(CD\) intersect on diameter at point \(E\). If \(\angle AEP = \angle DEP\) prove \(AB=CD\).
Let diameter's endpoint be \(P\). Given \(\angle AEP=\angle DEP\) suggests point \(E\) is equidistant angularly from arcs corresponding to \(AB\) and \(CD\). Construct perpendiculars from centre to both chords; using equal angles at \(E\) and properties of equal subtended angles and congruent triangles one can show half segments equal, hence \(AB=CD\). (Detailed triangle congruence proof follows from joining radii and using equal angles to get congruence.)
Q6 (figure 6.21). \(CD\) is diameter, \(CD \perp AB\) at \(E\). Show \(\triangle ABC\) is isosceles.
Since \(CD\) is diameter, \(C\) and \(D\) are endpoints of diameter; \(CD\perp AB\) at \(E\). Join \(AC\) and \(BC\). Note that \(CE\) is perpendicular bisector of chord \(AB\) (if \(E\) is midpoint). Then \(AE=BE\) and triangles \(AEC\) and \(BEC\) are congruent implying \(AC=BC\). Thus \(\triangle ABC\) is isosceles.

If you would like the remaining textbook practice problems solved step-by-step with neat MathJax diagrams (or downloadable PDF/print-friendly version), tell me and I'll prepare it in the exact format you prefer — directly below your website's menu without altering it.

Quick Reference & Tips

Key formula (relation between radius \(r\), distance \(d\) from centre to chord, and chord length \(l\))
\[ r^2 = d^2 + \left(\frac{l}{2}\right)^2 \] or equivalently \(l = 2\sqrt{r^2-d^2}\).
Remember
Perpendicular from centre to chord bisects it. Congruent chords ↔ equal distances from centre. Incircle centre = intersection of angle bisectors; circumcentre = intersection of perpendicular bisectors.
Designed with Comic Sans MS for comfortable mobile reading • Questions in red; answers in green • Math typeset with MathJax