Question 1: Fill in the blanks

  1. A line segment joining any point on the circle to its center is called a radius  of the circle.
  2. All the radii of a circle are equal.
  3. A line segment having its end points on a circle is called chord  of a circle.
  4. A chord that passes through the center of the circle is called a diameter  of the circle.
  5. Diameter of a circle is twice  its radius.
  6. A diameter is the largest  chord of the circle.
  7. The interior of a circle together with the circle is called the area of the circle.
  8. A chord of a circle divides the whole circular region into two parts, each called a  segment.
  9. Half of a circle is called a semicircle.
  10. A segment of a circle containing the center is called the major segment of the circle.
  11. The mid point of the diameter of a circle is the center  of the circle.
  12. The perimeter of the circle is called its circumference.

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 Question 2: State which of the following statements are true or false:

  1. Diameter of a circle is a part of a semi-circle of a circle : True
  2. Two semi-circles of a circle together make the whole circle: True
  3. Two semi-circular regions of a circle together make the whole circular region:  True
  4. An infinite number of chords may be drawn in a circle: True
  5. A line can meet a circle at the most at two points: True
  6. An infinite number of diameters can be drawn in a circle: True
  7. A circle has an infinite number of radii: True
  8. A circle consists of an infinite number of points: True
  9. Center of a circle lies on a circle: True

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Question 3: From an external point P, 29 cm away from the center of a circle, a tangent PT of length 21 cm is drawn. Find the radius of the circle.

Answer:

Radius = \sqrt{(29^2-21^2 )}= 20 cm  

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Question 4: Two tangents  PM \ and\  PN   are drawn from an exterior point P to a circle with center O . Prove that: \angle OPM \cong  \angle OPN   331

Answer:

In  \Delta PMO \ and\  \Delta PNO,  

PO is common, OM=ON (radius of the circle)  and

PN=PM   (Tangents to a circle from one point  circle are equal.)

Hence   \Delta PMO \cong  \Delta PNO  

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Question 5: In the given figure,  \Delta ABC   is inscribed in a circle with center O . If  \angle ACB=40^{\circ}   , find angle  332

Answer:

\angle ABC+ \angle BCA+ \angle BAC=180^{\circ}   

\angle BAC=90^o     (angle in a semicircle is a right angle).

\Rightarrow  \angle ABC=180-90-40=50^{\circ}     

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Question 6: In the given figure, O is the centre of a circle.  \Delta ABC   is inscribed in this circle. If  AB = AC, \ find\  \angle ABC \ and\  \angle ACB   .333

Answer:

\angle BAC=90^{\circ}     (angle in a semicircle is a right angle).

AB=AC  

\Rightarrow \angle CBA= \angle CBA=x^{\circ}   

\Rightarrow 2x+90=180^o  

\Rightarrow x=45^{\circ}   

\therefore \angle CBA= \angle CBA=45^{\circ}   

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Question 7: In the given figure, O is the center of a circle. If  \angle ABC = 54^{\circ}   , find \angle ACB   . Also, if \angle BCD = 43^{\circ}  , find \angle CBD   .p13

Answer:

\angle CBA+ \angle BAC+ \angle ACB=180^o  

\angle BAC=90^{\circ}     (angle in a semicircle is a right angle).

\Rightarrow  \angle ACB=180-90-54=36^{\circ}     

Similarly

\angle CBD+ \angle CDB+ \angle BCD=180^{\circ}     

\angle BDC=90^{\circ}      (angle in a semicircle is a right angle).

\Rightarrow  \angle CBD=180-90-43=47^{\circ}   

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Question 8: In the given figure, \Delta ABC   is inscribed in a circle with center O . If \angle ABC=(3x - 7)^{\circ}   \ and\  \angle ACB=(x -7)^{\circ}    , find the value of  x   .p14

Answer:

\angle CBA+ \angle BAC+ \angle ACB=180^o  

\angle BAC=90^{\circ}     (angle in a semicircle is a right angle).

\Rightarrow (3x-7)+90+(x+13)=180 \Rightarrow x=21^{\circ}     

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Question 9: In the adjoining figure, PRT is a tangent to the circle with center O . QR is a diameter of the circle. If  \angle QPR = 53^{\circ}   \ and\  \angle PQR = x^{\circ}    , then find the value of  x .p15

Answer:

\angle RPQ+ \angle PQR+ \angle QRP=180^o  

\angle QRP=90^{\circ}      (tangent is perpendicular to the line drawn from the center to the point of contact)

\Rightarrow \angle PQR=x=37^{\circ}      

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Question 10: In the adjoining figure, PRT is a tangent to the circle with center O . OR is the radius of the circle at the point of contact. P, O are joined and produced to the point Q on the circle. If  \angle RPO = 28^{\circ}  , \angle POR =x^{\circ}   \ and\  \angle ORQ =y^{\circ}    , then find the values of x and y   .p16

Answer:

\angle PRO=90    (tangent is perpendicular to the line drawn from the center to the point of contact)

\Rightarrow 90+x+28=180^o \Rightarrow x=62^{\circ}     

Also  2 \angle RQO= \angle ROP  

\Rightarrow \angle ORQ+\angle RQO+ \angle QOR=180^o  

\Rightarrow 31+118+y=180^o  

\Rightarrow y=31^{\circ}      

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Question 11: In the adjoining figure, PT is a tangent to the circle with center O , QT is a diameter of the circle. If  PT = QT \ and\  \angle QPT = x^{\circ}    , then find the value of x .p17

Answer:

Given  PT = QT  

\Rightarrow  \angle QPT= \angle PQT = x  

\Rightarrow 2x+90=180^o \Rightarrow x=45^{\circ}  

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Question 12: In the adjoining figure, PX and PY are tangents drawn from an exterior point P to a circle with center O and radius 8 cm. If PX = 15 \ cm, OP = a \ cm, PY = b \ cm ,  \angle POX = 56^{\circ}   \ and\  \angle OPY = x^{\circ}        then find the value of a,\  b\  and\  x .p18

Answer:

a= OP= \sqrt{(8^2+15^2 )}= 17 cm  

b= PY= \sqrt{(17^2-8^2 )}= 15 cm  

\angle POX = \angle POY=56^o  

\Rightarrow  56+90+x=180^o  

\Rightarrow x=34^{\circ}   

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Question 13: In the adjoining figure, AB is the diameter of the circle with center O . If  \angle ABM = 124^{\circ}   \ and\  CAB = x^{\circ}    , then find the value of x .p19

Answer:

180- 124+90+x=180^o  \Rightarrow  

x=34^{\circ}   

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Question 14: In the adjoining figure, PQ is a diameter of a circle with center O .  \Delta PQR   is an isosceles triangle with RP=RQ PQ is produced to a point S such that RQ = QS . If  \angle QPR = x \ and\  \angle QSR =y   , then find the values of x \ and\  y   .p20

Answer:

In  \Delta PQR, RP=RQ \\ \Rightarrow  \angle RPQ= \angle RQP=x  

And we know

\angle PRQ=90 \Rightarrow x=45^{\circ}   

\angle RQS=180-45=135^{\circ}   

Given  RQ=QS \Rightarrow \angle QRS=y  

\Rightarrow In \Delta PRS, 45+y+y+90=180^o \Rightarrow y=22 \frac{1}{2}^{\circ}     

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