Q1. What is a chord of a circle?

Ans: A segment with both endpoints on the circle (e.g. \(AB\)).

Q2. What is a diameter?

Ans: A chord passing through the centre; it equals \(2\times\) radius.

Q3. What is a central angle?

Ans: An angle with vertex at centre \(O\) subtending an arc (e.g. \(\angle AOB\)).

Q4. Define minor arc.

Ans: The shorter arc between two points on circle.

Q5. Define major arc.

Ans: The longer arc between two points on circle.

Q6. Semicircular arc measure (in degrees)?

Ans: \(180^\circ\).

Q7. If a perpendicular from centre meets chord \(AB\) at \(P\), what is \(AP\) compared to \(PB\)?

Ans: \(AP = PB\) (the chord is bisected).

Q8. If two chords are equal, what can you say about their arcs?

Ans: Their corresponding arcs are congruent (equal measures).

Q9. If \(\angle AOB = 60^\circ\), what is measure of minor arc \(AB\)?

Ans: \(m(\widehat{AB})=60^\circ\) (arc measure = central angle).

Q10. If arc \(AB\) is \(100^\circ\), what is reflex (major) arc \(AB\)?

Ans: Major arc measure \(=360^\circ-100^\circ=260^\circ\).

Q11. What is chord corresponding to semicircular arc?

Ans: A diameter.

Q12. If chord length is 7 cm and perpendicular from centre meets it at midpoint, what is half-chord length?

Ans: \(7/2=3.5\) cm.

Q13. If radius is \(10\) cm and distance from centre to chord is \(6\) cm, what is half chord length \(PB\)?

Ans: \(PB=\sqrt{10^2-6^2}=\sqrt{64}=8\) cm.

Q14. For a circle, what is the relation between equal central angles and arcs?

Ans: Equal central angles subtend congruent arcs.

Q15. If a diameter is perpendicular to a chord, what does it do to the chord?

Ans: It bisects the chord and its arc.

Q16. If a chord passes through the centre, what is it?

Ans: A diameter.

Q17. Minor or major: arc with measure \(200^\circ\)?

Ans: Major arc (since \(>180^\circ\)).

Q18. If central angle is \(90^\circ\), arc is called?

Ans: A quarter (or 90°) arc — a minor arc of \(90^\circ\).

Q19. If two arcs are congruent, what about their chords?

Ans: Their corresponding chords are congruent (equal length).

Q20. If arc \(DB\) corresponds to central angle \(\angle AOB\), state relation of measures.

Ans: \(m(\widehat{DB}) = m(\angle AOB)\).

Q1. In a circle centre \(O\), chord \(PQ = 7\) cm. If \(OP \perp PQ\) meets at \(A\). Find \(AP\).

Ans: Perpendicular from centre bisects chord ⇒ \(AP = PQ/2 = 3.5\) cm.

Q2. Radius \(=25\) cm; chord length \(48\) cm. Find distance of chord from centre.

Ans: Half chord \(=24\). If distance = \(d\), \(d=\sqrt{25^2-24^2}=\sqrt{625-576}=\sqrt{49}=7\) cm.

Q3. Radius \(=10\) cm; chord at distance \(9\) cm from centre has length ?

Ans: half-chord \(=\sqrt{10^2-9^2}=\sqrt{19}=4.3589\) cm ⇒ chord \(\approx 8.7178\) cm.

Q4. If diameter \(=24\) cm and chord \(=10\) cm, find distance from centre.

Ans: radius \(=12\). half-chord \(=5\). distance \(=\sqrt{12^2-5^2}=\sqrt{144-25}=\sqrt{119}\approx10.908\) cm.

Q5. If chord \(AB\) is bisected at \(P\) by radius \(OP\), and \(OP=6\), \(OB=10\). Find \(AB\).

Ans: \(PB=\sqrt{OB^2-OP^2}=\sqrt{100-36}=\sqrt{64}=8\) ⇒ \(AB=2\times8=16\) cm.

Q6. Two chords are equal; show that their subtended central angles are equal (short explanation).

Ans: Equal chords are at equal distance from centre; equal perpendiculars give equal right triangles ⇒ corresponding central angles equal.

Q7. Circle centre \(O\). If \(m(\widehat{AXD}) = 40^\circ\) (minor arc) then what is \(m(\angle AOD)\)?

Ans: \(m(\angle AOD)=40^\circ\) (central angle equals arc measure).

Q8. Diameters \(PQ\) and \(RS\) are perpendicular. Show arc \(PS \cong\) arc \(SQ\).

Ans: Perpendicular diameters split circle into four equal arcs \(=90^\circ\). So arcs \(PS\) and \(SQ\) both \(=90^\circ\) ⇒ congruent.

Q9. In circle centre \(O\), chord \(AB = CD\). What about arcs \(AB\) and \(CD\)?

Ans: arcs \(AB\) and \(CD\) are congruent (equal measure).

Q10. Central angles \(\angle AOB = \angle COD = 35^\circ\). Are arcs \(AB\) and \(CD\) congruent?

Ans: Yes — equal central angles subtend congruent arcs.

Q11. If minor arc \(AB = 60^\circ\), what is major arc \(AB\)?

Ans: \(360-60=300^\circ.\)

Q12. Show that perpendicular from centre to chord bisects the arc too (brief).

Ans: Perpendicular bisects chord ⇒ symmetric about line through centre ⇒ subtended angles equal ⇒ arcs equal, hence arc bisected.

Q13. In a circle, chord length doubles. What happens to distance from centre (qualitative)?

Ans: If chord longer (toward diameter), its perpendicular distance from centre decreases (approaches 0 for diameter).

Q14. Given radius \(r\) and distance \(d\) from centre to chord, chord length formula?

Ans: chord length \(=2\sqrt{r^2-d^2}.\)

Q15. If arc \(AXD\) is minor and arc \(AYB\) is major for chord \(AB\), which is smaller?

Ans: arc \(AXD\) is the minor (shorter) arc.

Q16. If central angle is \(120^\circ\), is arc minor or major?

Ans: It's a minor arc only if central angle \(<180^\circ\). So \(120^\circ\) gives a minor arc of \(120^\circ\).

Q17. Two equal arcs imply what about corresponding chords?

Ans: The corresponding chords are equal in length.

Q18. If chord AB is 16 cm and half-chord PB=8, find distance from centre (radius 10 cm)?

Ans: \(d=\sqrt{10^2-8^2}=\sqrt{100-64}=\sqrt{36}=6\) cm.

Q19. If central angle corresponding to arc DB is \(\angle AOB\), express \(m(\widehat{DAB})\).

Ans: \(m(\widehat{DAB}) = 360^\circ - m(\angle AOB)\) when arc DAB is the major arc opposite the central angle.

Q20. If central angle is 100° on one side and another central angle is 100° opposite (perpendicular diameters example), find the remaining angles if given in figure with symmetry.

Ans: The four sectors will be \(100^\circ,100^\circ,80^\circ,80^\circ\) accordingly (depends on figure details).

Q1. In circle centre \(O\), chord \(PQ=13\) cm; perpendicular from centre meets chord at \(Q\) (midpoint). Find \(QB\).

Ans: If chord length \(=13\), half-chord \(=6.5\). So \(QB=6.5\) cm (midpoint property).

Q2. Radius \(25\) cm, chord \(48\) cm — find distance from centre (worked).

Ans: half-chord \(=24\). distance \(d=\sqrt{25^2-24^2}=\sqrt{625-576}=\sqrt{49}=7\) cm. (Hence perpendicular from centre is 7 cm away.)

Q3. Radius \(=10\) cm chord at distance \(9\) cm from centre — length of chord (worked).

Ans: half-chord \(=\sqrt{10^2-9^2}=\sqrt{19}\). chord \(=2\sqrt{19}\approx8.7178\) cm.

Q4. Prove: perpendicular from centre to chord bisects the chord (sketch proof).

Ans: Let chord \(AB\), centre \(O\). Draw perpendicular \(OP\) to \(AB\). Triangles \(OAP\) and \(OBP\) are right, \(OA=OB\) (radii), and \(OP\) common ⇒ triangles congruent by RHS ⇒ \(AP=PB\).

Q5. Two equal chords \(AB\) and \(CD\) are at distances \(d_1\) and \(d_2\) from centre. Show \(d_1=d_2\).

Ans: Equal chords have equal perpendicular distances from centre (contraposition of earlier property). Use congruent right triangles to show equality.

Q6. Circle centre \(O\). \( \angle AOB = 100^\circ, \angle COD = 100^\circ\). Show arc \(AB \cong CD\) and chord \(AB\cong CD\).

Ans: Equal central angles ⇒ equal arcs. Equal arcs ⇒ equal corresponding chords. Hence arc \(AB\cong CD\) and chord \(AB\cong CD\).

Q7. Diameter \(MN\) in circle centre \(O\). If \(\angle MON = 180^\circ\) and \(\angle AOB=35^\circ\) in figure — find mirror angles (worked example from prompt).

Ans: For given figure if \(\angle AOB=35^\circ\) and opposite sector also 35°, then arcs equal and respective chords equal. (Refer to particular diagram to name results.)

Q8. If central angle is \(35^\circ\) and another central angle is \(35^\circ\), show minor arcs equal and list congruent chords (concrete statement).

Ans: minor arcs subtended by \(35^\circ\) are equal; corresponding chords equal. (List depending on labelling.)

Q9. Given radius \(=13\), chord length \(=10\), find distance from centre.

Ans: half-chord \(=5\). distance \(=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\) cm.

Q10. Show by calculation: circle radius \(=15\), chord at distance \(9\) ⇒ chord length?

Ans: half-chord \(=\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12\) ⇒ chord \(=24\) cm.

Q11. Two equal chords are at equal distance from centre. Prove they are equal (short argument).

Ans: Perpendiculars from centre give equal right triangles with equal hypotenuse (radii) and equal perpendiculars ⇒ halves of chords equal ⇒ chords equal.

Q12. If chord \(AB\) subtends \(60^\circ\) at centre, find major arc \(AB\) and minor arc numbers.

Ans: minor arc \(=60^\circ\), major arc \(=300^\circ\).

Q13. If chord \(AB\) is 16 cm, radius 10 cm, find distance from centre (worked).

Ans: half-chord \(=8\). distance \(=\sqrt{10^2-8^2}=\sqrt{36}=6\) cm.

Q14. If arc DB corresponds to \(\angle AOB\), express \(m(\widehat{DAB})\) (worked).

Ans: If \(\widehat{DB}\) = central angle \(\angle AOB\), then major arc \(DAB=360^\circ - m(\angle AOB)\).

Q15. In the prompt example: radius 10, distance 6 → chord length computed. Reproduce steps.

Ans: half-chord \(=\sqrt{10^2-6^2}=8\) ⇒ chord \(=16\) cm (as in solved Ex (2) in prompt).

Q16. If perpendicular from centre to chord meet at \(P\), how does it affect the arcs on either side?

Ans: It bisects both the chord and the corresponding minor arc (splits the arc into equal arcs).

Q17. Show that diameter perpendicular to chord bisects arc and chord (short proof provided earlier).

Ans: By congruent triangles as before; symmetry leads to arc bisected.

Q18. Given perpendicular distance 0 from centre (i.e., chord passes through centre). What is chord?

Ans: Chord passes through centre ⇒ chord is diameter ⇒ length \(=2r\).

Q19. Numerical: chord 13, radius 10 — impossible? explain.

Ans: chord cannot exceed diameter \(=20\). 13 is OK. If chord \(>20\) impossible. Check that \(half\ chord =6.5\) and \(distance=\sqrt{10^2-6.5^2}\) real.

Q20. Application: A circle has radius \(r\). If chord length equals \(r\), find distance from centre.

Ans: half-chord \(=r/2\). distance \(d=\sqrt{r^2-(r/2)^2}=\sqrt{r^2 - r^2/4}=\sqrt{3r^2/4}=\tfrac{\sqrt{3}}{2}r.\)

Q1. In a circle centre \(P\), chord \(AB=13\) cm. \(PQ\perp AB\) at \(Q\). Find \(QB\).

Solution: Perpendicular from centre bisects chord ⇒ \(QB = AB/2 = 13/2 = 6.5\) cm.

Q2. Radius \(=25\) cm. Find distance of chord from centre if chord length is \(48\) cm.

Solution: half-chord \(=24\). distance \(d=\sqrt{25^2-24^2}=\sqrt{625-576}=\sqrt{49}=7\) cm.

Q3. O is centre. Chord length \(24\) cm at distance \(9\) cm from centre. Find radius.

Solution: half-chord \(=12\). radius \(= \sqrt{d^2 + 12^2} = \sqrt{9^2 + 12^2} = \sqrt{81+144}=\sqrt{225}=15\) cm.

Q4. Centre \(C\), radius \(=10\) cm, chord \(=12\) cm. Find distance from centre.

Solution: half-chord \(=6\). distance \(d=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\) cm.

Q1. Diameters \(PQ\) and \(RS\) perpendicular at centre \(C\). Explain why arc \(PS\) and arc \(SQ\) are congruent and list other congruent arcs.

Solution: Perpendicular diameters divide circle into four equal quadrants each \(=90^\circ\). So arc \(PS=90^\circ\) and arc \(SQ=90^\circ\) ⇒ congruent. Other congruent arcs: \(PR\), \(RQ\) etc — any opposite quadrant arcs equal.

Q2. In figure with diameter \(MN\) and central angles given (example from prompt): (1) find \(m\angle AOB\) and \(m\angle COD\); (2) show arc \(AB\cong CD\); (3) show chord \(AB\cong CD\).

Solution sketch: If given central angles are \(35^\circ\) and \(35^\circ\), then \(m\angle AOB=35^\circ\) and \(m\angle COD=35^\circ\). Hence arcs \(AB\) and \(CD\) both \(35^\circ\) ⇒ congruent. Congruent arcs ⇒ corresponding chords equal: \(AB=CD.\)