CHAPTER 9 – CIRCLES (Class 9, NCERT/CBSE)
1-Mark (20)
2-Mark (20)
3-Mark (20)
Exercise 9.1 – Solutions
Exercise 9.2 – Solutions
Exercise 9.3 – Solutions
Part A — 1-Mark Questions (20) with Solutions
Q1. Define a circle.
The set of all points in a plane at a fixed distance (radius) from a fixed point (centre).
Q2. What is a chord? What is the longest chord?
A line segment joining two points on a circle. The longest chord is a diameter.
Q3. What is meant by “angle subtended by a chord at the centre”?
The angle formed at the centre by the two radii drawn to the endpoints of the chord.
Q4. State Theorem 9.1 in short.
Equal chords subtend equal angles at the centre.
Q5. State the converse (Theorem 9.2).
Chords that subtend equal central angles are equal.
Q6. What does the perpendicular from the centre to a chord do?
It bisects the chord (Theorem 9.3).
Q7. If a line from the centre bisects a chord, then…?
It is perpendicular to the chord (Theorem 9.4).
Q8. Equal chords are ______ from the centre.
Equidistant (Theorem 9.5).
Q9. Chords equidistant from the centre are ______.
Equal (Theorem 9.6).
Q10. The angle at the centre is how many times the angle at the circumference on the same arc?
Twice (Theorem 9.7): \(\angle \text{centre} = 2\angle \text{circumference}\).
Q11. Angles in the same segment are…?
Equal (Theorem 9.8).
Q12. Angle in a semicircle equals…?
\(\,90^\circ\).
Q13. What is a cyclic quadrilateral?
A quadrilateral whose all vertices lie on a circle.
Q14. Opposite angles of a cyclic quadrilateral are…?
Supplementary: each pair sums to \(180^\circ\) (Theorem 9.10).
Q15. State Theorem 9.11 (converse).
If a pair of opposite angles of a quadrilateral is \(180^\circ\), it is cyclic.
Q16. If a chord subtends \(\,120^\circ\) at the centre, what is the angle in the corresponding segment?
\(\,60^\circ\).
Q17. Two equal chords of a circle are at distances \(4\text{ cm}\) and \(4\text{ cm}\) from the centre. Are they equal?
Yes, equal distance \(\Rightarrow\) equal chords (Th. 9.6).
Q18. The radius is perpendicular to a chord at its midpoint. True/False?
True (Th. 9.3 and 9.4 together).
Q19. If chord \(AB= \text{radius } R\), what is \(\angle AOB\) at the centre?
Using \(AB=2R\sin(\tfrac{\theta}{2})=R\Rightarrow \sin(\tfrac{\theta}{2})=\tfrac12\Rightarrow \theta=60^\circ\).
Q20. If \(\angle APB=40^\circ\) (P on circle), find \(\angle AOB\).
\(\angle AOB=2\angle APB=80^\circ\).
Part B — 2-Mark Questions (20) with Solutions
Q1. Prove: Equal chords subtend equal angles at the centre.
Let \(AB=CD\) in circle with centre \(O\). In \(\triangle AOB\) and \(\triangle COD\): \(OA=OC\), \(OB=OD\) (radii), \(AB=CD\). By SSS, triangles congruent \(\Rightarrow \angle AOB=\angle COD\).
Q2. Prove: If chords subtend equal central angles, then the chords are equal.
If \(\angle AOB=\angle COD\), then by SAS in \(\triangle AOB\) and \(\triangle COD\) (two radii equal and included angles equal), triangles are congruent \(\Rightarrow AB=CD\).
Q3. Show: The perpendicular from the centre to a chord bisects the chord.
Let \(OM\perp AB\). In right \(\triangle OMA\) and \(\triangle OMB\): \(OA=OB\) (radii), \(OM\) common, and right angles at \(M\). RHS \(\Rightarrow AM=MB\).
Q4. Show the converse: If a line through the centre bisects a chord, it is perpendicular to it.
Let \(M\) be midpoint of \(AB\) and \(O\) centre. In \(\triangle OMA\) and \(\triangle OMB\): \(OA=OB\), \(AM=MB\), \(OM\) common. SSS \(\Rightarrow \angle OMA=\angle OMB\). They form a linear pair, so each is \(90^\circ\). Hence \(OM\perp AB\).
Q5. Prove: Equal chords are equidistant from the centre.
For equal chords \(AB,CD\), with perpendiculars \(OM,ON\) to them, right triangles \(\triangle OMA\) and \(\triangle ONC\) are congruent (RHS). Hence \(OM=ON\).
Q6. Prove the converse: Chords equidistant from the centre are equal.
If \(OM=ON\) and \(OM\perp AB\), \(ON\perp CD\), then right triangles \(\triangle OMA\) and \(\triangle ONC\) are congruent (RHS). Hence \(AB=CD\).
Q7. Find the central angle if the angle at the circumference is \(37^\circ\) on the same arc.
\(\angle \text{centre}=2\times 37^\circ=74^\circ\).
Q8. Find the angle in the segment if the central angle is \(150^\circ\).
\(\angle \text{segment}=\dfrac{150^\circ}{2}=75^\circ\).
Q9. In a circle, chord \(AB\) subtends \(80^\circ\) at the centre. Find \(\angle ACB\) for any \(C\) on the circle not on arc \(AB\).
\(\angle ACB=\dfrac{80^\circ}{2}=40^\circ\).
Q10. If \(AB\) is a diameter and \(C\) is any point on the circle, show \(\angle ACB=90^\circ\).
Diameter subtends \(180^\circ\) at the centre, hence angle at the circumference is \(\tfrac{180^\circ}{2}=90^\circ\).
Q11. Two chords \(AB\) and \(CD\) intersect at \(E\) inside a circle and \(\angle AEB=120^\circ\). If \(\angle ADB=40^\circ\), find \(\angle ACB\).
Angles in same segment: \(\angle ACB=\angle ADB=40^\circ\).
Q12. A chord \(AB\) is at distance \(d\) from centre \(O\) in circle radius \(R\). Write its length.
\(AB=2\sqrt{R^{2}-d^{2}}\).
Q13. If equal chords \(AB\) and \(CD\) subtend angles \(\alpha\) and \(\beta\) at a point \(P\) on the circle (same segment), compare \(\alpha\) and \(\beta\).
They are equal, since both equal chords subtend equal central angles and hence equal angles at any circumference point in same segment.
Q14. In a cyclic quadrilateral \(ABCD\), if \(\angle A=78^\circ\), find \(\angle C\).
\(\angle A+\angle C=180^\circ\Rightarrow \angle C=102^\circ\).
Q15. In a circle, chords \(AB\) and \(CD\) are equal. Prove arcs \(AB\) and \(CD\) are congruent.
Equal chords subtend equal central angles \(\Rightarrow\) corresponding arcs have equal measure \(\Rightarrow\) congruent arcs.
Q16. If two arcs are congruent, what can be said about their corresponding chords?
They are equal.
Q17. In a circle, \(\angle ABC=35^\circ\) and points \(A,B,C,D\) are concyclic. Find \(\angle ADC\).
Angles in the same segment: \(\angle ADC=\angle ABC=35^\circ\).
Q18. In an isosceles trapezium, show it is cyclic.
Non-parallel sides equal \(\Rightarrow\) base angles equal \(\Rightarrow\) a pair of adjacent angles are supplementary \(\Rightarrow\) cyclic (Th. 9.11).
Q19. If two circles have radii \(5\text{ cm}\) and \(5\text{ cm}\), show they are congruent.
Same radius \(\Rightarrow\) congruent circles by definition.
Q20. If \(AB\) is a chord and \(O\) is the centre, express \(\angle AOB\) in terms of arc \(AB\).
\(\angle AOB=\) measure of arc \(AB\) (in degrees) at centre; any inscribed angle on arc \(AB\) equals \(\tfrac12\angle AOB\).
Part C — 3-Mark Questions (20) with Solutions
Q1. Prove that the angle in a semicircle is a right angle.
Let diameter \(AB\) subtend \(\angle ACB\) at \(C\). Central angle \(\angle AOB=180^\circ\Rightarrow \angle ACB=\dfrac{180^\circ}{2}=90^\circ\).
Q2. Show that equal chords are equidistant from the centre, and conversely.
Drop perpendiculars from the centre to both chords and use RHS congruence both ways (Theorems 9.5 and 9.6).
Q3. In a circle, prove that angles in the same segment are equal.
Let arc \(PQ\) subtend \(\angle POQ\) at centre and \(\angle PAQ,\angle PCQ\) at circumference. By Th. 9.7, \(\angle PAQ=\tfrac12\angle POQ=\angle PCQ\).
Q4. If a line segment subtends equal angles at two points on the same side of it, prove the four points are concyclic (Theorem 9.9).
Construct the circle through three of the points; the fourth must lie on it because the same chord subtends equal inscribed angles only on the circle.
Q5. In a circle, show that the longer chord is nearer to the centre.
Let chords \(AB\) and \(CD\) with \(AB>CD\). Distances \(OM\) and \(ON\) to chords satisfy \(AB=2\sqrt{R^2-OM^2}\), \(CD=2\sqrt{R^2-ON^2}\). From \(AB>CD\Rightarrow OM
Q6. Prove that the perpendicular bisectors of all chords of a circle pass through the centre.
From Theorem 9.4, the line through the centre that bisects a chord is perpendicular. The unique common intersection for different chords is the centre.
Q7. In a circle, chord \(AB\) equals radius \(R\). Find angles subtended by \(AB\) at (i) centre, (ii) any point on minor arc, (iii) any point on major arc.
(i) \(AB=2R\sin(\tfrac{\theta}{2})=R\Rightarrow \theta=60^\circ\). (ii) \(\tfrac{\theta}{2}=30^\circ\). (iii) \(\tfrac{360^\circ-\theta}{2}=150^\circ\).
Q8. Prove: Opposite angles of a cyclic quadrilateral are supplementary.
Let \(\angle A\) subtend arc \(BC\) and \(\angle C\) subtend arc \(AB\). Then \(\angle A=\tfrac12 \widehat{BC}\), \(\angle C=\tfrac12 \widehat{AB}\) and \(\widehat{AB}+\widehat{BC}=360^\circ\Rightarrow \angle A+\angle C=180^\circ\).
Q9. Two circles intersect at \(A\) and \(B\). Through \(B\), draw tangents \(BP\) and \(BQ\) to the circles. Prove \(P,A,Q\) are collinear.
Using equal tangent theorem, \(\angle PBA = 90^\circ - \angle BCA\) and similarly for the other circle; the pair of right complements implies \(\angle PBQ\) subtends a straight line through \(A\).
Q10. Two equal chords \(AB\) and \(CD\) intersect at \(E\). Prove the segments are equal: \(AE=DE\) and \(BE=CE\).
Equal chords subtend equal angles at the centre and thus equal angles at circumference; using vertical-opposite angles at \(E\) and AAS congruence of the triangles formed by the intersecting chords yields corresponding segments equal.
Q11. In \(\triangle ABC\), if circles with diameters \(AB\) and \(AC\) meet at \(A\) and \(P\), prove \(P\) lies on \(BC\).
Since \(\angle APB=90^\circ\) and \(\angle APC=90^\circ\), the right angles imply \(B,P,C\) are collinear (unique right angle from \(A\)).
Q12. In a cyclic parallelogram, prove it is a rectangle.
In a parallelogram, opposite angles are equal; in a cyclic quadrilateral, opposite angles are supplementary. Hence each must be \(90^\circ\Rightarrow\) rectangle.
Q13. In a circle with centre \(O\), chords \(AB\) and \(CD\) are parallel with \(AB>CD\). Compare \(OM\) and \(ON\) (perpendicular distances).
Longer chord is nearer the centre: \(OM
Q14. Show: If \(\angle A=\angle C\) in quadrilateral \(ABCD\), then \(ABCD\) is cyclic iff \(\angle B=\angle D\).
If cyclic, opposite angles are supplementary, hence equal pairs force equality of the other pair. Converse: if both pairs equal, sums are \(180^\circ\Rightarrow\) cyclic.
Q15. In a circle, equal arcs subtend equal chords and equal angles at the centre—prove any one implies the others.
Use central angle–arc measure equivalence and Theorems 9.1 & 9.2 to cycle the implications.
Q16. If \(\angle APB=40^\circ\) and \(\angle AQB=50^\circ\) with \(P,Q\) on circle, find \(\angle AOB\).
\(\angle AOB=2\angle APB=80^\circ\). (Note: the inscribed angle is independent of the point on the same arc.)
Q17. In circle radius \(R\), a chord subtends \(2\alpha\) at the centre. Find its distance from the centre.
Half-chord \(=R\sin\alpha\). Distance \(d=\sqrt{R^2-(R\sin\alpha)^2}=R\cos\alpha\).
Q18. In cyclic \(ABCD\), if \(\angle ABC=72^\circ\) and \(\angle BAD=38^\circ\), find \(\angle BCD\).
\(\angle BCD=180^\circ-\angle BAD=142^\circ\).
Q19. Prove: If a chord through the midpoint of another chord is perpendicular to it, then the first chord is a diameter.
Perpendicular from centre bisects a chord. If a chord through the midpoint is perpendicular, it must pass through centre \(\Rightarrow\) it’s a diameter.
Q20. For a triangle inscribed in a circle radius \(R\), side \(a\) opposite angle \(A\) satisfies \(a=2R\sin A\). State without proof.
It follows from chord–angle formula: chord \(=2R\sin(\tfrac{\text{central angle}}{2})=2R\sin A\).
Exercise 9.1 — Textbook Questions & Perfect Solutions
| Question | Solution |
|---|---|
| Q1. Two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. | Let radii \(R\) be equal and chords \(AB\) and \(CD\) be equal in the two circles with centres \(O,O'\). In \(\triangle AOB\) and \(\triangle CO'D\): \(OA=O'A'=R\), \(OB=O'D=R\), \(AB=CD\). By SSS, triangles are congruent \(\Rightarrow \angle AOB=\angle CO'D\). |
| Q2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. | If \(\angle AOB=\angle CO'D\) and \(OA=O'A'=R\), \(OB=O'D=R\), then by SAS, \(\triangle AOB\cong \triangle CO'D\). Hence \(AB=CD\). |
Exercise 9.2 — Textbook Questions & Perfect Solutions
| Question | Solution |
|---|---|
| Q1. Two circles of radii \(5\text{ cm}\) and \(3\text{ cm}\) intersect at two points and the distance between their centres is \(4\text{ cm}\). Find the length of the common chord. | Let centres be \(O_1,O_2\), radii \(r_1=5, r_2=3\), distance \(d=4\). Distance from \(O_1\) to common chord along \(O_1O_2\): \(x=\dfrac{r_1^2-r_2^2+d^2}{2d}=\dfrac{25-9+16}{8}=4\). Half-chord \(=\sqrt{r_1^2-x^2}=\sqrt{25-16}=3\). Hence chord \(=2\times 3=\boxed{6\text{ cm}}.\) |
| Q2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. | Let equal chords \(AB\) and \(CD\) intersect at \(E\). Angles \(\angle AEB\) and \(\angle CED\) are vertical. Also, equal chords \(\Rightarrow\) equal arcs, hence \(\angle EAB=\angle ECD\) (same segment). By AAS, \(\triangle AEB\cong \triangle CED\)\(\Rightarrow AE=CE\) and \(BE=DE\). |
| Q3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. | With notation of Q2 and centre \(O\), drop perpendiculars \(OM\perp AB\), \(ON\perp CD\). Equal chords \(\Rightarrow OM=ON\) (equidistant). In right triangles \(\triangle OME\) and \(\triangle ONE\): \(OM=ON\), \(OE\) common, and \(\angle OME=\angle ONE=90^\circ\). RHS \(\Rightarrow \triangle OME\cong \triangle ONE\). Thus \(\angle (EO,EA)=\angle (EO,EC)\); i.e., \(OE\) makes equal angles with both chords. |
| Q4. If a line intersects two concentric circles with centre \(O\) at \(A, B, C, D\) (in that order), prove that \(AB=CD\). | The configuration is symmetric about \(O\). Perpendicular distance of the line from \(O\) is same for both sides. The intercept outside the inner circle on one side equals that on the other side. By equal right triangles formed with radii at perpendiculars (or by symmetry), \(\boxed{AB=CD}\). |
| Q5. On a circle of radius \(5\text{ m}\), three girls stand at equal distances on the boundary. Distances \(RS=SM=6\text{ m}\). Find \(RM\). | Let central angle between adjacent girls be \(\theta\). Chord \(6=2R\sin\frac{\theta}{2}=10\sin\frac{\theta}{2}\Rightarrow \sin\frac{\theta}{2}=0.6\). Then \(\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=2(0.6)(0.8)=0.96\). Required chord \(RM=2R\sin\theta=10\times 0.96=\boxed{9.6\text{ m}}.\) |
| Q6. A circular park of radius \(20\text{ m}\) has three boys at equal distances on the boundary. Find the length of each string if they talk pairwise by toy telephones. | Points are equally spaced \(\Rightarrow\) central angle \(=120^\circ\). Chord length \(=2R\sin(60^\circ)=2\cdot 20\cdot \dfrac{\sqrt{3}}{2}=\boxed{20\sqrt{3}\text{ m}}\) (each string). |
Exercise 9.3 — Textbook Questions & Perfect Solutions
| Question | Solution |
|---|---|
| Q1. In a circle with centre \(O\), \(\angle BOC=30^\circ\) and \(\angle AOB=60^\circ\). If \(D\) is any point on the circle other than arc \(ABC\), find \(\angle ADC\). | \(\angle AOC=\angle AOB+\angle BOC=90^\circ\). Angle in the same segment: \(\angle ADC=\dfrac{\angle AOC}{2}=45^\circ\). |
| Q2. A chord equals the radius. Find the angle subtended by the chord at a point on (i) minor arc, (ii) major arc. | Let central angle be \(\theta\). \(AB=R=2R\sin(\tfrac{\theta}{2})\Rightarrow \theta=60^\circ\). (i) On minor arc: \(\dfrac{\theta}{2}=30^\circ\). (ii) On major arc: \(\dfrac{360^\circ-\theta}{2}=150^\circ\). |
| Q3. In \(\triangle PQR\) on a circle, \(\angle PQR=100^\circ\). Find \(\angle OPR\) where \(O\) is centre. | Inscribed angle \(100^\circ\) intercepts arc \(PR\) of \(200^\circ\). The smaller central angle is \(360^\circ-200^\circ=160^\circ\Rightarrow \angle POR=160^\circ\). In isosceles \(\triangle POR\), base angles \(=\dfrac{180^\circ-160^\circ}{2}=10^\circ\). Hence \(\angle OPR=\boxed{10^\circ}\). |
| Q4. In \(\triangle ABC\), \(\angle ABC=69^\circ\), \(\angle ACB=31^\circ\). Find \(\angle BDC\) with \(D\) on the circle. | \(\angle BAC=180^\circ-69^\circ-31^\circ=80^\circ\). Angles in same segment: \(\angle BDC=\angle BAC=\boxed{80^\circ}\). |
| Q5. Points \(A,B,C,D\) on a circle; \(AC\) and \(BD\) meet at \(E\). If \(\angle BEC=130^\circ\) and \(\angle ECD=20^\circ\), find \(\angle BAC\). | Note \(\angle ECD=\angle ACD=20^\circ\Rightarrow\) arc \(AD=40^\circ\). Also, angle between intersecting chords: \(\angle BEC=\tfrac12(\widehat{BC}+\widehat{AD})\Rightarrow 130^\circ=\tfrac12(\widehat{BC}+40^\circ)\Rightarrow \widehat{BC}=220^\circ\). Hence minor arc \(BC=360^\circ-220^\circ=140^\circ\Rightarrow \angle BAC=\tfrac12(140^\circ)=\boxed{70^\circ}\). |
| Q6. \(ABCD\) cyclic, diagonals meet at \(E\). If \(\angle DBC=70^\circ\) and \(\angle BAC=30^\circ\), find \(\angle BCD\). Further, if \(AB=BC\), find \(\angle ECD\). | \(\angle DBC=\angle DAC=70^\circ\) (same segment). Then at \(A\): \(\angle BAD=\angle BAC+\angle CAD=30^\circ+70^\circ=100^\circ\). In cyclic quadrilateral, \(\angle BCD=180^\circ-\angle BAD=80^\circ\). If \(AB=BC\), in \(\triangle ABC\): base angles at \(A\) and \(C\) are \(30^\circ\). Thus \(\angle ACD=\angle BCD-\angle BCA=80^\circ-30^\circ=50^\circ\). Since \(E\) lies on \(AC\), \(\angle ECD=\boxed{50^\circ}\). |
| Q7. If diagonals of a cyclic quadrilateral are diameters of the circumcircle, prove the quadrilateral is a rectangle. | Each diagonal subtends \(180^\circ\) at centre \(\Rightarrow\) any angle standing on a diameter is \(90^\circ\). All four angles right angles \(\Rightarrow\) rectangle. |
| Q8. If the non-parallel sides of a trapezium are equal, prove it is cyclic. | Let \(AB\parallel CD\) and \(AD=BC\). Base angles at \(A\) and \(B\) are equal; likewise at \(D\) and \(C\). Then \(\angle A+\angle C=180^\circ\Rightarrow\) cyclic (Theorem 9.11). |
| Q9. Two circles intersect at \(B\) and \(C\). Through \(B\), chords \(ABD\) and \(PBQ\) are drawn meeting the circles at \(A,D\) and \(P,Q\) respectively. Prove \(\angle ACP=\angle QCD\). | Note \(\angle ABP\) and \(\angle ADQ\) stand on the same arc \(AB\) and \(DQ\) respectively in their circles. Using equal angles in the same segment transferred through point \(C\) and parallel chord directions, one gets \(\angle ACP\) equals \(\angle QCD\). |
| Q10. If circles are drawn with two sides of a triangle as diameters, prove that their second point of intersection (other than the common vertex) lies on the third side. | Let diameters on \(AB\) and \(AC\) meet again at \(P\). Then \(\angle APB=90^\circ\) and \(\angle APC=90^\circ\). Hence \(B,P,C\) are collinear, i.e., \(P\) lies on \(BC\). |
| Q11. \(ABC\) and \(ADC\) are right triangles with common hypotenuse \(AC\). Prove \(\angle CAD=\angle CBD\). | Both right triangles lie on the same circle with diameter \(AC\). Thus \(B\) and \(D\) are concyclic with \(A,C\). Angles \(\angle CAD\) and \(\angle CBD\) stand on the same arc \(CD\) (or \(CB\)) \(\Rightarrow\) equal. |
| Q12. Prove that a cyclic parallelogram is a rectangle. | In a parallelogram, opposite angles are equal; in a cyclic quadrilateral, opposite angles are supplementary. Hence each equals \(90^\circ\Rightarrow\) rectangle. |
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