**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.