Circle – Maharashtra Board Class 9 Solutions for Geometry
AlgebraGeometryScience and TechnologyHindi
Exercise – 4.1
Solution 1:
Points in the interior: O, A, B, C.
Points in the exterior: E, D, G, F.
Points on the circle: P, Q, T, R.
Solution 2:
The radius of the circle is 6.7 cm.
d(P, R) = 5.7 cm.
∴5.7 cm < 6.7 cm.
∴ The distance between P and R is less than the radius of the circle.
∴Point R lies in the interior of the circle.
d(P, Q) = 7.6 cm.
∴7.6 cm> 6.7 cm.
∴ The distance between P and Q is greater than the radius of the circle.
∴Point Q lies in the exterior of the circle.
Solution 3:
Only one common chord (joining the points of intersection) can be drawn between two common points of two intersecting circles.
Solution 4:
Radius (r) = 7cm …(Given)
OP=4cm
∴4 < 7
∴OP < r
∴ Point P is in the interior of the circle.
OQ=9cm
∴ 9 > 7
∴OQ > r
∴ Point Q is on the exterior of the circle.
Solution 5:
Let the two circles with centre O and P intersect each other at the points A and B.
AB is the common chord.
Line l passes through the centre O and P.
Draw seg OA, seg OB, seg PA, seg PB.
Let line l intersect chord AB in point M.
In ∆AOP and ∆BOP,
seg OA ≅ seg OB … (Radii of the same circle)
seg PA ≅ seg PB … (Radii of the same circle)
seg OP ≅ seg OP … (Common side)
∴ ∆ AOP ≅ ∆BOP … (sss test)
∴∠AOP ≅ ∠BOP … (c.a.c.t.)
i.e. ∠AOM ≅ ∠BOM
In ∆AOM and ∆BOM
seg OA ≅ seg OB … (Radii of the same circle)
∠AOM ≅ ∠BOM … (Proved)
seg OM ≅ seg OM … (Common side)
∴ ∆AOM ≅ ∆BOM… (SAS test)
∴seg AM ≅ seg BM …(c.s.c.t.)
and ∠AMO ≅ ∠BMO …(c.a.c.t.)
But m∠AMO + m∠BMO = 180° … (Angles in a linear pair)
∴m∠AMO = m∠BMO=90°
∴seg OP is the perpendicular bisector of seg AB.
i.e. line l is the perpendicular bisector of the common chord AB.
Solution 6:
Solution 7:
In ∆AOB and ∆COD,
seg OA ≅ seg OD; seg OB ≅ seg OC … (Radii of the same circle)
∠AOB ≅ ∠COD … (Vertically opposite angles)
∴ ∆AOB ≅ ∆COD … (SAS test)
∴ seg AB ≅ seg CD … (c.s.c.t.)
i.e. chord AB ≅ chord CD.
Solution 8:
Solution 9:
Join OQ and OR.
In ∆OPQ and ∆OPR
seg OQ ≅ seg OR … (Radii of the same circle)
seg PQ ≅ seg PR … (Given)
seg PO ≅ seg PO … (Common side)
∴ ∆OPQ ≅ ∆OPR … (SSS test)
∴∠OPQ ≅ ∠OPR … (c.a.c.t.)
∴seg PO is the bisector of ∠QPR.
i.e. the bisector of ∠RPQ passes through the centre of the circle.
Solution 10:
Solution 11:
Solution 12:
Solution 13:
Solution 14:
Draw seg AM ⊥ chord PS.
The perpendicular drawn from the centre of a circle to a chord bisects the chord.
∴In the larger circle, PM = MS …(1)
and in the smaller circle, QM = MR …(2)
Now, PM – QM=PQ … (P-Q-M) …(3)
AndMS – MR=RS … (M-R-S) …(4)
From (1), (2), (3) and (4), PQ = RS.
Solution 15:
Circles with centres O and P intersect each other in points Q and R
Seg QR is the common chord.
Seg OP intersects the common chord QR in point S.
OQ = OR … (Radii of the same circle)
∴Point O is equidistant from points Q and R. …(1)
Similarly, PQ=PR
∴Point P is equidistant from points Q and R … (2)
From (1) and (2) and by Perpendicular Bisector Theorem, OP is the perpendicular bisector of QR.
∴point S is the midpoint of QR and
m∠QSO = m∠QSP = 90° …(3)
QR=12cm …(Given)…(4)
From (3) and (4), QS = 6cm …(5)
In right angled ∆PQS,
By Pythagoras’ Theorem
PQ2 = QS2 + PS2
∴(8)2 = (6)2 + PS2…[Given and from (5)]
∴PS2 = 82 – 62 = 64 -36
∴PS2 = 28
∴PS= √28
∴PS= 2√7 cm …(6)
Similarly, in right angled ∆OQS,
QO2 = QS2 +OS2
∴ (10) 2=(6) 2+OS2 …[Given and from (5)]
∴OS2 = 10 2 – 6 2 = 100 – 36
∴OS2 =64
∴OS= 8cm …(7)
OP = OS +SP …[O – S – P]
= (8+2√7) cm
The distance between their centres is (8+2√7) cm
Solution 16:
The perpendicular drawn from the centre of a circle to a chord bisects the chord.
∴Point E is the midpoint of chord AB.
∴ seg AE ≅ seg BE … (1)
Diameter CD ⊥ chord AB … (Given)
In ∆CAE and ∆CBE,
seg AE ≅ seg BE … [From (1)]
∠CEA ≅ ∠CEB
seg CE ≅ seg CE … (Common side)
∴ ∆CAE≅ ∆CBE … (SAS test)
∴seg CA ≅ seg CB … ( c.s.c.t.)
i.e. CA= CB
∴ ∆ABC is an isosceles triangle.