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.

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: $\angle OPM \cong \angle OPN$

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

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

$\angle BAC=90^o$  (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  $AB = AC, \ find\ \angle ABC \ and\ \angle ACB$.

$\angle BAC=90^{\circ}$  (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 $\angle CBD$.

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

$\angle BAC=90^{\circ}$  (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  $x$.

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

$\angle BAC=90^{\circ}$ (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  $x$.

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

$\angle QRP=90^{\circ}$   (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 $x and y$.

$\angle PRO=90$  (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 $x$.

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 $a,\ b\ and\ x$.

$a= OP= \sqrt{(8^2+15^2 )}= 17 cm$

$b= PY= \sqrt{(17^2-8^2 )}= 15 cm$

$\angle POX = \angle POY=56^o$

$\Rightarrow 56+90+x=180^o$

$\Rightarrow x=34^{\circ}$

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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 $x$.

$180- 124+90+x=180^o \Rightarrow$

$x=34^{\circ}$

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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 $x \ and\ y$.

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  $RQ=QS \Rightarrow \angle QRS=y$

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