| Book Name | : Ganit Prakash |
| Subject | : Mathematics |
| Class | : 10 (Madhyamik) |
| Publisher | : Prof Nabanita Chatterjee |
| Chapter Name | : Theorems Related To Circle (3rd Chapter) |
LET US WORK OUT – 3.1
Question 1
Let us see the adjoining figure of the circle with centre O and write the radii which are situated in the segment PAQ.
Solution :
In the circle with centre O,OP, OA, OC, OQ are the radii in the segment PAQ.
Question 2
Let us write in the following by understanting it.
(i) In a circle there are number of points.
Solution :
Infinite.
(ii) The greatest chord of the circle is
Solution :
Diameter.
(iii) The chord divides the circular region into two
Solution :
Sector.
(iv) All diameters of the circle pass through
Solution :
Centre.
(v) If two segments are equal, then their two arcs are in length.
Solution :
Equal.
(vi) The sector of the circular region is the region enclosed by the arc and the two
Solution :
Radii.
(vii) The length of the line segment joining the point outside the circle and the centre is than the length of radius.
Solution :
Greater.
Question 3
With the help of scale and pencil, compass let us draw a circle and indicate centre, chord diameter, radius, major arc, minor arc on it.
Solution :
Centre – O
Chord – CD
Diameter – AB
Radius – OE = OA = OB
Minor arc - \overline{x y}
Major arc - \overline{x p y}
4. Let us write true or false :
Question 1
The circle is a plane figure.
Solution :
[True]
Circle is a rectilinear figure
Question 2
The segment is a plane region.
Solution :
[True]
Segment of a circle is rectilinear figure.
Question 2
The sector is a plane region.
Solution :
[True]
Sector is a rectilinear.
Question 3
The chord is a line segment.
Solution :
[True]
Chord is a line segment.
Question 4
The arc is a line segment.
Solution :
[True]
Arc is a line segment.
Question 5
There are a finite number of chords of same length in a circle.
Solution :
[False]
A circle contains infinite number of equal chords.
Question 6
One and only one circle can be drawn by taking a fixed point as its centre.
Solution :
[False]
Only one circle can be drawn with a centre.
Question 7
The lengths of the radii of two congruent circles are equal.
Solution :
[True]
Lengths of radius of two congruent circles are equal.
Application 3.
I draw a chord P Q of the circle with the centre A, which is not a diameter. I draw a perpendicular AM on PQ from A. I prove with reason that PM = MQ. [Let me prove it by drawing]
Solution :
PQ is a chord of a circular with centre A. AM is the perpendicular from A to PQ. To prove, P M=M Q Join A, P and A, Q.
Proof : In two right angled triangles APM AQM, Hypotenuse AP = Hypotenuse AQ [Radii of same circle] AM common.
\therefore \triangle A P M \cong \triangle A Q M (R.H.S)
\therefore PM = MQ (Coresponding sides). Proved.
Application 4.
I prove the theorem-33 by the proof of congruency of \triangle O A D \text{ and }\triangle OBD with the help of S-A-S axiom of congruency. [Let me do it myself].
Solution :
In \triangle O A D \ \triangle O B D
O A=O B \text { (radii of same circle) }
\angle O A D=\angle O B D (Corresponding angle)
AD = DB [\because D is the mid point of AB]
\therefore \triangle O A D \cong \triangle O B D
\therefore \angle \mathrm{ODA}=\angle \mathrm{ODB} (Corresponding angle)
The adjacent angles formed by OD standing on AB are equal.
\therefore OD is perpendicular on AB.
Application 5.
The perpendicular distance of a chord from the centre of a circle, having the radius of 17 cm is 8 cm in length. Let us write by calculating the length of its chord. [Let me do it myself]
Solution :
Here OB = 17cm, OD = 8cm.
\therefore B D^2=17^2-8^2= 289-64 = 225
\therefore \mathrm{BD}=\sqrt{225}=15 cm
\therefore \text{ Chord } \mathrm{AB}=2 \times 15, cm=30 cm.
Application 6.
In a circle with the radius of 5cm in length, the two parallel chords of length 8cm and 6cm are situated on opposite sides of the centre. Let us write by calculating the distance between two chords. [Let me do it myself]
Solution :
ABCD are two parallel chords of circle with centre O.
Chords are opposite sides of the centre O.
Here, \mathrm{ PB =\frac{1}{2} \ of \ 8 cm=4 cm \ QD=\frac{1}{2} \ of \ 6 cm=3 cm \ O B=O D=5 cm. }
\mathrm{ \therefore O P^2=O B^2-P B^2=5^2-4^2=25-16=9 }
\therefore OP = 3 cm.
& \mathrm{ OQ^2=O D^2-Q D^2=5^2-3^2=25-9=16 }
\therefore OQ = 4cm
\therefore P Q=(4+3) \mathrm{cm} = 7 cm.
\therefore Distance between the chords = 7 cm.
Application 8.
I prove with reason that two equal chords of any circle are equidistant from its centre.
Solution :
To prove that two chords of a circle equidistance from the centre are equal. Let there be two chords AB & CD of the circle with centre O.
And O E \perp A B, O F \perp C D O E=O F
To prove, AB = CD.
Proof : In two right angled \triangle A O E \triangle C O F, hypotenuse OA = OC (radii of same circle)
\mathrm{OE}=\mathrm{OF} \text { (given) }
\therefore \triangle A O E \cong \triangle C O F
\therefore AE = CF (Corresponding sides)
\therefore 2 \times A E=2 \times C F
\therefore A B=C D \text { Proved. }
Application 9.
Let us prove that the perpendicular bisector of a chord of a circle passes through its centre. [Let me prove it myself]
Solution :
The prove that the perpendicular of any chord is passing through the centre of the circle.
Let O is the centre and AB is a chord of the circle.
O D \perp A B, if possible, let the perpendicular biscet meets at O1.
\therefore O D \perp A B \ and \ O^1 D \perp A B.
i.e., OD & D1 both are perpendicular on AB; it will be possible it OD° and OD1 coincide with each other.
\therefore The perpendicular bisector of the chord is passing through the centre.
Application 10.
Let us prove that a straight line can not intersect a circle at more than two points. [Let me prove it myself]
Solution :
To prove that a straight line can not cut a circle at not more than two points.
Ans. If possible let the straight line cuts two circles with centre O, at A, B C points.
Draw the perpendicular on OD from the centre O.
\therefore AD = DB & AD = DC.
\therefore \mathrm{DB}=\mathrm{DC}; it will be possible if points B coincide with point C.
\therefore A straight line can not cut a circle at more than two points.
LET US WORK OUT - 3.2
Application 1
The length of a radius of a circle with its centre O is 5cm and the length of its chord AB is 8 cm. Let us write by calculating the distance of the chord AB from the centre O.
Solution :
Centre O, chord AB = 8cm.
Radius = OA = 5cm. AD =4 cm.
O D^2=O A^2-A D^2 \\
=(5)^2-(4)^2=25-16
\mathrm{OD}^2=9 . \therefore \mathrm{OD}=3 cm
Application 2
The length of a diameter of a circle with its centre at O is 26 cm. The distance of the chord PQ from the point O is 5cm. Let us write by calculating the length of the chord PQ.
Solution :
Diameter = 26cm
\therefore \text { Radius }=13 cm
\therefore OP = 13cm
OD = 5cm.
P D^2=O P^2-O D^2 \\
=13^2-5^2=169-25=144
P D=\sqrt{144}=12 \\
P Q=2 \times P D=2 \times 12 cm .=24 cm.
Application 3
The length of a chord P Q of a circle with its centre O is 4 cm and the distance of from the point O is 2.1 cm. Let us write by calculating, the length of its diameter.
Solution :
Centre O; chord PQ =4 cm; OD = 2.1 cm.
P O^2=O D^2+P^2 \\
=(2.1)^2+(2)^2=4.41+4=8.41 \\
P O=\sqrt{8.41}=2.9
\therefore Radius = 2.9 cm; Diameter = 2 × 2.9 cm = 5.8 cm.
Application 4
The lengths of two chords of a circle with its centre at O are 6cm and 8cm. If the distance of the smaller chord from centre is 4cm, then let us write by calculating, the distance of the other chord from the centre.
Solution :
Centre O.
Larger chord CD = 8cm.
Small chord AB = 6cm
OP = 4cm ; AP = 3cm
\mathrm{ \therefore \text { In } \triangle A O P }
\mathrm{ O A^2=O P^2+A P^2=4^2+3^2=16+9=25 }
\mathrm{ O A=\sqrt{25}=5 cm }. \\
\mathrm{ \therefore O C=5 cm, C Q=\frac{1}{2} C D=\frac{1}{2} \times 8 cm=4 cm \\ }
\mathrm{ O Q^2=O C^2-C Q^2=5^2-4^2=25-16=9 \\ }
\mathrm{ O Q=\sqrt{9}=3 cm . }
Application 5
If the length of a chord of a circle is 48cm and the distance of it from the centre is 7cm, then let us write by calculating the length of radius of the circle.
Solution :
Centre O; AB = 48cm, OE = 7cm.
\mathrm{O A^2=A E^2+O E^2(24)^2+(7)^2 \\ }
\mathrm{=576+49=625 \\ }
\mathrm{\therefore O A=\sqrt{625}=25 \\ }
\mathrm{\therefore O C=25 cm ; O F=20 cm . \\ }
\mathrm{C F^2=O C^2-O F^2=25^2-20^2=625-400=225 \\ }
\mathrm{C F=\sqrt{225}=15 cm . \\ }
\mathrm{C D=2 \times C F=2 \times 15=30 cm }.
Length of the 2nd chord =30 cm.
Application 6
In the circle of adjoining figure with its centre at O, O P \perp A B; if A B=6 cm and PC = 2cm, then let us write by calculating the length of radius of the circle:
Solution :
Centre of the circle O.
Chord AB = 6cm, AP = 3cm, PC = 2cm.
Let Radius OA = OC = r cm.
\mathrm{O P=O C-P C=(r-2) \mathrm{cm} . \\ }
\mathrm{O A^2=O P^2+A P^2 \\ }
\mathrm{r^2=(r-2)^2+3^2 \\ }
\mathrm{r^2=r^2-4 r+4+9 \\ }
4r = 13. \therefore r=\frac{13}{4} cm .=3.25 cm .
Application 7
A straight line intersects one of the two concentric circles at the points A and B and the other at the points C and D. I prove with reason that AC = DB.
Solution :
A straight line AB cuts two concentric circles with centre O at A & B and C & D points.
To prove, AC = BD.
As OP is the perpendicular on the chord AB.
\therefore AP = BP
Again, OP is the perpendicular on the chord CD.
\therefore CP = DP
\therefore AP - CP = BP - DP
or, AC = BD (Proved)
Application 8
I prove that the two intersecting chords of any circle can not bisect each other unless both of them are diameters of the circle.
Solution :
Let O is centre of the circle. Two chords AB & CD interest each other at P such that the point p is the mid point of AB :
To prove : P is not the mid point of CD. Join O, P.
Proof : As P is the mid point by AB.
\because OP \perp AB
Again, chord AB & CD passing through P. It is not possible that AB & CD both are perpendicular on OP at P. As AB is perpendicular on OP.
\therefore CD is not the perpendicular on OP.
Again, the straight line joining the centre and mid point of the chord, is the perpendicula on the chord.
\therefore P is not the mid point of the chord CD.
If the both straight lines be the diameter, then they will bisect each other.
Application 9
The two circles with centres X and Y intersect each other at the points A and BA is joined with the mid-point ' S ' of XY and the perpendicular on SA through th point A is drawn which intersects the two circles at the points P and Q. Let us prov that PA = AQ.
Solution :
Two circles with centres X & Y intersect each other at A and B. S is the m point of the line joining X & Y. Join AS. The perpendicular on AS cut the circles at P & C To prove : AP = AQ.
Construction : From two centres X & Y, two perpendiculars drawn on PQ are XD & Y
Proof : As three straight lines XD, SA & YE are each perpendicular on PQ.
\therefore X D \| S A|| Y E
As S is the mid point of XY.
\therefore SX = SY
As three parallel straight lines XD, SA & YE cut the straight line X Y in two equal parts, then the lines XD, SA, YE cut the other straight PQ into two equal parts.
i.e., DA = AE[ As the straight line XD from X, will perpendicular on chord PA.]
\mathrm{\therefore DA = \frac{1}{2} AP }
Similarly, \mathrm{YE \perp A W, A E=\frac{1}{2} A Q }
as DA = AE \mathrm{\therefore \frac{1}{2} A P=\frac{1}{2} A Q \\ }
\therefore AP = AQ.
Application 10
The two parallel chords AB and CD with the lengths of 10 cm and 24 cm in a circle are situated on the opposite sides of the centre. If the distance between two chords AB and CD is 17 cm, then let us write by calculating, the length of the radius of the circle.
Hints : Let the length of the radius of the circle be r cm and the distance of the chord A B from the centre be x cm. \therefore The distance of the chord CD from the centre is (17-x)cm. \therefore r^2=x^2+5^2 \& r^2=(17-x)^2+(12)^2, So, x^2+5^2=(17-x)^2+12^2 \therefore x = 12.
Solution :
AB = 10cm \therefore AP = 5cm
Radius OA = OC = r cm. (Let)
CD = 24cm, CQ = 12cm
Let OP = x cm and OQ = (17 - x)( As OP + OQ = 17cm)
O C^2=O Q^2+C Q^2
r^2=(17-x)^2+12^2
Again \mathrm{OA}^2=\mathrm{OP}^2+\mathrm{AP}^2
r^2=x^2+5^2
x^2+25=(17-x)^2+12^2
x^2+25=289-34 x+x^2+144
34 x = 289 + 144 - 25
34 x = 408 \quad \therefore x=\frac{408}{34}=12
\therefore r^2=x^2+5^2 \\
=12^2+5^2 \\
= 144 + 25 = 169
\therefore r=\sqrt{169}=13
\therefore Radius of the circle = 13cm.
Application 11
The centres of two circles are P and Q; they intersect at the points A and B. The straight line parallel to the line segment PQ through the point A intersects the two circles at the points C and D. I prove that, CD = 2 PQ.
Solution :
From the centres P & Q two perpendiculars PE & QF are drawn on CD.
QF bisects AD ; AF=\frac{1}{2} AD
Again, PE bisects AC; AE=\frac{1}{2} C
EF = AF + AE
=\frac{1}{2} A D+\frac{1}{2} A O \\
EF=\frac{1}{2} CD
Again, PQ = EF
\therefore P Q=\frac{1}{2} C D \\
\therefore CD = 2 PQ (Proved)
Application 12
The two chords AB and AC of a circle are equal. I prove that the bisector of \angle B A C passes through the centre.
Solution :
AB and AC are two equal chords of the circle with centre O. Join OA and OB.
In \triangle A O C \ \& \ \triangle A O B
(i) OB = OC (Radii of same circle)
(ii) AO common
(iii) AB = AC (given)
\therefore \triangle A O C \cong \triangle A O B \\
\therefore \angle O A C=\angle O A B
\therefore OA is the bisector of \angle B A C.
\therefore Bisector of \angle B A C, is passing through centre (A).
Application 13
If the angle-bisector of two intersecting chords of a circle passes through it centre, then let me prove that the two chords are equal.
Solution :
Let AB & CD are two chords. They intersect at ' P. Join OA, OB, OC, OD. OP is the bisector of \angle B O C.
To prove, AB = CD.
\triangle C O P \cong \triangle B O P as (i) OP common,
(ii) OC = OB (Same radius) and (iii) \angle O P C=\angle O P B \ as \ O P is the bisector of \angle B O C.
\angle A O P=\angle C P O+\angle A P C \\
\angle D O P=\angle D P B+\angle B P D \\
\because \angle C P O=\angle B P O \\
& \angle A P C=\angle D P B (vertically opposite)
\therefore \angle A P O=\angle D P O
Now, in \triangle A O P \ \& \ \triangle D O P,
(i) OP is common, (ii) OA = OB (same radius) and (iii) \angle A P C=\angle D P B.
\therefore \triangle A O P \cong \triangle D O P \\
\because AP = PD
CP + PD = PB + AP
or, CD = AB (Proved).
Application 14
I prove that, among two chords of a circle, the length of the chord nearer centre is greater than the length of the other.
Solution :
AB & CD are two chords of the circle with centre O.
Their distances from the centre are OP & OQ.
OP \angle O Q. To prove AB > CD
Join OB & OD. As P is the midpoint of AB
\therefore P B=\frac{1}{2} AB
and Q is the mid point of CD.
\therefore Q D=\frac{1}{2} CD
\text { In } \triangle O P B, O B=\sqrt{P B^2+O P^2} \\
\text { In } \triangle O Q D, O D=\sqrt{O Q^2+Q D^2} \\
\therefore \sqrt{P B^2+O P^2}=\sqrt{O Q^2+Q D^2} \\
P B^2+O P^2=O Q^2+Q^2 \\
As O P<O Q \quad \therefore P B>Q D \\
\frac{1}{2} A B>\frac{1}{2} C D \\
AB > CD (Proved).
Application 15
Let us write by proving the chord with the least length through any point in a circle.
Solution :
P is a point inside the circle.
Chord AB passing through P will be smaller if the point P will be the midpoint of AB.
If P in the mid point of AB
\therefore O P \perp A B
As perpendicular distance is the shortest distance.
\therefore AB is smallest (Proved).
M.C.Q.
Question 1
The lengths of two chords of a circle with centre O are equal. If \angle A O B=60, then the value of \angle C O D is
(a) 40°
(b) 30°
(c) 60°
(d) 90°
Solution :
AB & CD are two equal chords of the circle with O if \angle A O B=60°, then \angle C O D= 60°.................... (c)
Question 2
The length of the radius of a circle is 13 cm and the length of a chord of a circle is 10 cm, the distance of the chord from the centre of the circle is
(a) 12.5 cm
(b) 12 cm
(c) \sqrt{69} cm
(d) 24 cm
Solution :
Radius = 13 cm. length of chord = 10 cm, Distance from the centre =\sqrt{13^2-5^2} =\sqrt{169-25}=\sqrt{144}=12 ........................ (b)
Question 3
AB and CD are two equal chords of a circle with its centre O. If the distance of the chord AB from the point O is 4 cm then the distance of the chord from the centre 0 of the circle is
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
Solution :
AB & CD are two equal chords.
If the distance of AB from the centre O = 4 cm.
then the distance of CD from the centre O = 4 cm ........... (b)
Question 4
The length of each of two parallel chords is 16 cm. If the length of the radius of the circle is 10 cm., then the distance between two chords is
(a) 12 cm.
(b) 16 cm.
(c) 20 cm.
(d) 5 cm
Solution :
Two parallel chords of equal length 16 cm.
Length of the radius = 10 cm.
Distance between them =\sqrt{10^2-8^2}+\sqrt{10^2-8^2}=6+6=12 cm. ........... (a)
Question 5
The centre of two concentric circles is O; a straight line intersects a circle at the points A and B and the other circle at the point C and D. If AC = 5 cm, then the length of BD is
(a) 2.5 cm
(b) 5 cm
(c) 10 cm
(d) None of these
Solution :
When a straight line cuts two concentric circles at A & B and C & D points, AC = BD.
BD = 5 cm .................. (b)
Let us write True/False :
Question 1
Only one circle can be drawn through three collinear points.
Solution :
False
Question 2
The two circles ABCDA and ABCEA are same circle.
Solution :
True
Question 3
If two circles AB and AC of a circle with its centre O are situated on the opposite sides of the radius OA, then \angle O A B=\angle O A C.
Solution :
False
Let us fill in the blanks :
Question 1
If the ratio of two chords PQ and RS of a circle with its centre O is 1 : 1, then, \angle P O Q : \angle R O S = _______________.
Solution :
\angle P O Q : \angle R O S = = 1 : 1
Question 2
The perpendicular bisector of any chord of a circle is _____________ of that circle.
Solution :
Passing through the centre.
Short Answers:
Question 1
Two equal circles of radius 10 cm intersect each other and the length of their common chord is 12 cm. Let us determine the distance between the two centres of the two circles.
Solution :
CD = 12 cm . \quad \therefore C O=\frac{12}{2}=6 cm.
AO =\sqrt{10^2-6^2}=\sqrt{64}=8 \\
AB = 2 × 8 = 16 cm.
\therefore Distance between two centres = 16 cm.
Question 2
AB and AC are two equal chords of a circle having the radius of 5 cm. The centre of the circle is situated at the outside of the traingle ABC. IF AB = AC = 6 cm, then let us calculate the length of the chord BC.
Solution :
AB = AC = 6 cm.
OA = 5 cm.
Let AP = a cm.
PO = b cm.
OA = a + b =5 cm.
BP = CP = n
a^2+n^2=6^2 \& b^2+n^2=5^2
a^2+n^2-b^2-n^2=6^2-5^2=36-25=11
a^2-b^2=11,(a-b) \times(a+b)=11 \quad \therefore(a-b) \times 5=11 a-b=\frac{11}{5} \\
a=\frac{1}{2} \cdot(\frac{11}{2}+5)=\frac{18}{5} \quad \therefore n=\sqrt{6^2-(\frac{18}{5})^2}=\sqrt{36-\frac{324}{25}}=\sqrt{\frac{576}{25}}=\frac{24}{5} \\
\therefore BC = 2n = 2 × \frac{24}{5} = 9.6 cm.
Question 3
The length of two chords AB and CD of a circle with its centre O are equal. If \angle A O B=60^{\circ} and CD = 6 cm, then let us calculate the length of the radius of the circle.
Solution :
AB = CD = 6 cm.
\angle A O B=60^{\circ}
\therefore \triangle A O B is an equilateral triangle.
\mathrm{OA}=\mathrm{OB}= 6 cm.
\therefore Radius = 6 cm.
Question 4
P is any point in a circle with its centre O. If the length of the radius is 5 cm and OP = 3 cm, then let us determine the least length of the chord passing through the point P.
Solution :
BP =\sqrt{5^2-3^2}
=\sqrt{25-9}=\sqrt{16}=4 \\
AB = 2 × BP = 2 × 4 = 8 cm.
Question 5
The two circles with their centres at P and Q intersect each other at the points A and B. Through the point A, a straight line parallel to PQ intersects the two circles at the points C and D respectively. If PQ = 5 cm, then let us determine the length of CD.
Solution :
PQ = 5 cm, CD = ?
CD = 2 PQ = 2 × 5 cm
= 10 cm .
