 In a circle whose radius is 8cm., chord is drawn at a point 3cm. from the centre of the circle. The chord is divided into two segments by a point on it. If one segment of the chord is 9cm .What is the length of the other segment?
Solution:
From the data given,
Radius, OB = 8 cm
OC = 3cm
AB is a chord, The chord is divided into two line segments AP and BP , let BP = 9 cm. Now we have to find AP,
To find AP:
Proof:
In Δ OCB ; ∟OCB = 90˚[since OC⊥AB]
BC^{2} + OC^{2} = OB^{2}
BQ^{2} = OB^{2} – OC^{2}
BQ = √(8^{2} – 3^{2}) = √(649) = √55
AB = 2√55
AP = AB – BP = (2√55 – 9)cm
 Suppose two chords of a circle are equidistant from the center of the circle. Prove that the chords have equal length.
Solution:
Figure is drawn by the data given in the problem. Let O be the centre of the circle. AB and CD are 2 chords which are equidistant from the centre O of the circle, since OB = OD
Now we have to prove OB = OD
Proof:
In ΔMBO and ΔNDO
∟BMO = ∟DNO = 90˚
∟BMO = ∟DNO = 90˚
MO = NO (data given)
OB = OD (radii)
ΔMBO = ΔNDO (RHS)
MB = DN
Hence, AB = CD
 Suppose two chords of a circle are unequal in length. Prove that the chord of larger length is nearer to the centre than the chord of smaller length
Solution:
In the figure, O is the center of the circle, where AB and CD are chords. Ab is larger and CD is smaller chord. OP and OQ the distance from the chord AB and CD respectively.
To prove: OP < OQ
Proof:
In ΔAPO; AQ^{2} = AP^{2} + PQ^{2} …………(1)
In ΔCQO; OC^{2} = CQ^{2} + OQ^{2} ………….(2)
AO = OC radius of the circle.
From (1) and (2),
AP^{2} + PO^{2} = CQ^{2} + OQ^{2}
Let AP = CQ + x [AB < CQ; ^{1}/_{2} AB > ^{1}/_{2} CD ; AP > CQ]
(CO+X)^{2} + PO^{2} = CQ^{2} + OQ^{2}
CQ^{2} + 2. CQ. X + PO^{2} = CQ^{2} + OQ^{2}
OQ^{2} = OP^{2} + 2.CQ.X
or
OQ^{2} > OP^{2}
OQ > OP
OP < OQ
 Let AB and CD be parallel chords of a circle with center O. M is the midpoint of AB and N is the midpoint of CD . Prove that O, N ,M are collinear. If MN = 3cm , AB= 4cm, CD = 10 cm find the radius of the circle.
Solution:
Given: AB and CD are to parallel chords. M and N are midpoint of AB and CD respectively.
To prove: O, N, M are collinear
To find: radius of the circle i.e, OD
Proof:
∟OMB = 90˚ (M is the mid point)
∟ONB = 90˚ and ∟MND = 90˚
∟OND + ∟MND = 180˚
Therefore, O, N , M are collinear.
To find the radius of the circle with centre O.
Let ON = x, in ΔOMB
OB^{2} = OM^{2} + MB^{2}
= (3 + x)^{2} + 2^{2 }(AB = 4cm and MB = 2 cm)
= 9 + 6x + x^{2} + 4
OB^{2} = x^{2} + 6x + 13 ……(1)
In ΔOND, OD^{2} = ON^{2} + ND^{2}
= x^{2} + 5^{2} (CD = 10cm and ND = 5 cm)
OD^{2} = x^{2} + 25
OB^{2} = x^{2} + 25 …………(2) (OB = OD)[CD = 10cm and ND = 5cm)
x^{2} + 6x + 13 = x^{2}+ 5^{2}
6x = 25 – 13 = 12
x = ^{12}/_{6} = 2
Thus, x = 2 cm
OD^{2} = x^{2} + 5^{2}
OD^{2} = 2^{2} + 5^{2}
= 4 + 25 = 29
OD = √29
 In the figure ,ABCD is a straight line two concentric circles with centre O. Is it possible to have CD = 6cm, OD = 9cm and OB = 5 cm. Justify.
Solution:
Let us solve the problem using the Pythagoras theorem.
In the figure it is given that OD = 9 cm, OB = 5cm and CD = 6 cm.
Constrction: Draw a line OP from O perpendicular to AD, which cuts the chord AD at P.
Therefore, we have ΔOPD where ∟OPD = 90 ˚, Thus, ΔOPD is a right angled triangle.
By Pythagoras theorem, OD^{2} = OP^{2}+ PD^{2}
We know OD = 9 thus, 9^{2}= OP^{2}+ PD^{2}, where 9^{2} is not a Pythagorian Number.
Therefore, CD = 6cm, OD = 9cm and OB = 5 cm is not possible.
In the fig A, C, B are collinear centres of three circles and AC =CB. Prove that PQ = RS where the line PQRS passes through the point of intersection Q and R and cuts other two circles at P and Q.
[Hint: draw perpendiculars from A, C, B to PQ]
Solution:
Statement  Reason 
AC = CB  Given

Draw AD, CE and BF perpendicular to PQ

Construction 
ADCEBF  corresponding angles are equal i.e., 90˚ 
DE = EF  equal intercepts 
DQ + QE = ER + RF

⊥ from the centre bisects the chord 
^{1}/_{2}PQ + QE = ER + ^{1}/_{2}RS

QE = ER 
^{1}/_{2}PQ = ^{1}/_{2}RS


PQ = RS
^{ } 
 Suppose AB and CD are two parallel chords in a circle .prove that the line joining this midpoint passes through the centre of the circle.
Solution:
Given : O center of the circle and AB CD
Construction : Join OM and ON
Proof: M is the midpoint of CD
∟OMB =90 ˚
Similarly N is midpoint of CD
∟OND =90 ˚
∟OMB + ∟BMN = 180 ˚ (∟BMN =90 ˚)
i.e OMN is a straight line
MN passes through O.
 Prove that a parallelogram inscribed in a circle is a rectangle
Solution:
Parallelogram ABCD is cyclic
∟A + ∟C = 180 ˚ [opposite angle of cyclic quadrilateral]
But ∟A = ∟C = 90 ˚ ……….(1) [opposite angle of parallelogram are equal]
Similarly, ∟B + ∟D = 180˚
∟B = ∟D = 90˚ ………………..(2) [opposite angle of parallelogram are equal]
AB = CD and AD = BC ………..(3) [Opposite sides of the parallelogram]
Opposite sides are equal and each angle is right angle therefore, ABCD is rectangle.
1 thought on “Circles – Exercise 4.4.2 – Class IX”
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