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Question 1: From a point $Q$, the length of the tangent to a circle is 24 cm and the distance of $Q$ from the centre is 25 cm. The radius of the circle is:
Solution: (A) 7 cm
Step 1: Let radius be $r$. By Pythagoras theorem in $\Delta OPQ$ (where $P$ is point of contact), $OQ^2 = OP^2 + PQ^2$.
Step 2: $25^2 = r^2 + 24^2 \Rightarrow 625 = r^2 + 576$.
Step 3: $r^2 = 625 - 576 = 49 \Rightarrow r = 7$ cm. -
Question 2: If $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $\angle POQ = 110^\circ$, then $\angle PTQ$ is equal to:
Solution: (B) $70^\circ$
Step 1: In quadrilateral $POQT$, sum of angles is $360^\circ$. $\angle OPT = \angle OQT = 90^\circ$.
Step 2: $\angle PTQ + \angle POQ = 180^\circ$.
Step 3: $\angle PTQ = 180^\circ - 110^\circ = 70^\circ$. -
Question 3: If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^\circ$, then $\angle POA$ is equal to:
Solution: (A) $50^\circ$
Step 1: $\angle APB = 80^\circ$. Line $OP$ bisects $\angle APB$, so $\angle APO = 40^\circ$.
Step 2: In $\Delta OAP$, $\angle OAP = 90^\circ$.
Step 3: $\angle POA = 180^\circ - (90^\circ + 40^\circ) = 50^\circ$. -
Question 4: The length of a tangent from a point $A$ at distance 5 cm from the centre of the circle is 4 cm. The radius of the circle is:
Solution: (A) 3 cm
Step 1: Let radius be $r$. Distance $d=5$, Tangent $l=4$.
Step 2: $r = \sqrt{d^2 - l^2} = \sqrt{5^2 - 4^2}$.
Step 3: $r = \sqrt{25 - 16} = \sqrt{9} = 3$ cm. -
Question 5: Two concentric circles are of radii 5 cm and 3 cm. The length of the chord of the larger circle which touches the smaller circle is:
Solution: (C) 8 cm
Step 1: The chord is tangent to the smaller circle. The radius of smaller circle is perpendicular to the chord and bisects it.
Step 2: Half length of chord $= \sqrt{5^2 - 3^2} = \sqrt{25-9} = \sqrt{16} = 4$ cm.
Step 3: Total length $= 2 \times 4 = 8$ cm. -
Question 6: A tangent $PQ$ at a point $P$ of a circle of radius 5 cm meets a line through the centre $O$ at a point $Q$ so that $OQ = 12$ cm. Length $PQ$ is:
Solution: (D) $\sqrt{119}$ cm
Step 1: In right $\Delta OPQ$, $OQ^2 = OP^2 + PQ^2$.
Step 2: $12^2 = 5^2 + PQ^2 \Rightarrow 144 = 25 + PQ^2$.
Step 3: $PQ^2 = 144 - 25 = 119 \Rightarrow PQ = \sqrt{119}$ cm. -
Question 7: From a point $P$ which is at a distance of 13 cm from the centre $O$ of a circle of radius 5 cm, the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is:
Solution: (A) 60 cm$^2$
Step 1: Length of tangent $PQ = \sqrt{13^2 - 5^2} = \sqrt{169-25} = 12$ cm.
Step 2: Area of $\Delta POQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30$ cm$^2$.
Step 3: Area of quadrilateral $PQOR = 2 \times \text{Area}(\Delta POQ) = 2 \times 30 = 60$ cm$^2$. -
Question 8: At one end $A$ of a diameter $AB$ of a circle of radius 5 cm, tangent $XAY$ is drawn to the circle. The length of the chord $CD$ parallel to $XY$ and at a distance 8 cm from $A$ is:
Solution: (D) 8 cm
Step 1: Distance of chord from centre $O$ is $8 - 5 = 3$ cm.
Step 2: Half length of chord $= \sqrt{5^2 - 3^2} = \sqrt{25-9} = 4$ cm.
Step 3: Total length $= 2 \times 4 = 8$ cm. -
Question 9: If $O$ is the centre of a circle, $PQ$ is a chord and the tangent $PR$ at $P$ makes an angle of $50^\circ$ with $PQ$, then $\angle POQ$ is equal to:
Solution: (A) $100^\circ$
Step 1: $\angle OPQ = 90^\circ - 50^\circ = 40^\circ$.
Step 2: In $\Delta OPQ$, $OP=OQ$ (radii), so $\angle OQP = 40^\circ$.
Step 3: $\angle POQ = 180^\circ - (40^\circ + 40^\circ) = 100^\circ$. -
Question 10: If two tangents inclined at an angle $60^\circ$ are drawn to a circle of radius 3 cm, then length of each tangent is equal to:
Solution: (D) $3\sqrt{3}$ cm
Step 1: The line joining the external point to the centre bisects the angle between tangents. So angle becomes $30^\circ$.
Step 2: In the right triangle formed, $\tan 30^\circ = \frac{\text{radius}}{\text{tangent}} = \frac{3}{L}$.
Step 3: $\frac{1}{\sqrt{3}} = \frac{3}{L} \Rightarrow L = 3\sqrt{3}$ cm.