NeokaalStudy

Circles: Practice Worksheet

Tangent lengths, tangent angles, and theorem-based reasoning with circles.

18 problems·35–45 min·★★★★★
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  1. Problem 1
    Tangent Length·★★☆☆☆
    From an external point T, tangent TA touches a circle with centre O. If $OT=14$ cm and $\angle OTA=30^\circ$, find TA.
    1. (A)$7\sqrt{2}$ cm
    2. (B)$7\sqrt{3}$ cm
    3. (C)$14\sqrt{3}$ cm
    4. (D)$7$ cm
    ▶Answer
    $7\sqrt{3}$ cm
    ▶Step-by-step solution
    1. Since OA is a radius to the point of contact, $OA\perp TA$.
    2. In right triangle OTA, $\cos 30^\circ=TA/14$.
    3. Therefore $TA=14\times\frac{\sqrt{3}}{2}=7\sqrt{3}$ cm.
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  2. Problem 2
    Tangent Length·★★★☆☆
    Two tangents from P to a circle of radius 7 cm make an angle of 90°. Find the length of each tangent.
    1. (A)7 cm
    2. (B)14 cm
    3. (C)$7\sqrt{2}$ cm
    4. (D)$7/2$ cm
    ▶Answer
    7 cm
    ▶Step-by-step solution
    1. The line joining P to the centre bisects the angle between the equal tangents.
    2. Use one right triangle: its angle at P is $90/2=45^\circ$, with radius opposite and tangent adjacent.
    3. Thus $\tan 45^\circ=\text{radius}/\text{tangent}$, giving tangent length 7 cm.
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  3. Problem 3
    Tangent Length·★★★☆☆
    A point P is 39 cm from the centre O of a circle of radius 36 cm. Tangents PA and PB touch the circle. Find the area of quadrilateral PAOB.
    1. (A)270 cm²
    2. (B)585 cm²
    3. (C)540 cm²
    4. (D)1404 cm²
    ▶Answer
    540 cm²
    ▶Step-by-step solution
    1. A radius is perpendicular to the tangent at its point of contact.
    2. In right triangle OAP, $PA=\sqrt{39^2-36^2}=15$ cm.
    3. The quadrilateral consists of two right triangles, so its area is $2\times\frac12\times 36\times 15=540$ cm².
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  4. Problem 4
    Angle Between Tangents·★★★☆☆
    Tangents at A and B to a circle meet at P. If $\angle AOB=145^\circ$, find $\angle APB$.
    1. (A)18°
    2. (B)35°
    3. (C)72°
    4. (D)145°
    ▶Answer
    35°
    ▶Step-by-step solution
    1. The radii OA and OB are perpendicular to the tangents PA and PB.
    2. Thus quadrilateral OAPB has two right angles.
    3. So $\angle APB=180^\circ-145^\circ=35^\circ$.
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  5. Problem 5
    Angle Between Tangents·★★★★☆
    Tangents PA and PB meet at P, and $\angle APB=100^\circ$. Find $\angle OAB$, where O is the centre.
    1. (A)80°
    2. (B)50°
    3. (C)30°
    4. (D)40°
    ▶Answer
    40°
    ▶Step-by-step solution
    1. The equal tangents PA and PB make the configuration symmetric about OP.
    2. The angle between radii OA and OB is $180^\circ-100^\circ=80^\circ$.
    3. Triangle AOB is isosceles, so each base angle is $40^\circ$.
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  6. Problem 6
    Angle Between Tangents·★★★☆☆
    Assess this claim as True or False: chord AB makes a 60° central angle, so the two tangents through A and B must also enclose 60°.
    ▶Answer
    False
    ▶Step-by-step solution
    1. Each radius is perpendicular to its tangent.
    2. The central angle and the angle between the tangents are supplementary.
    3. The tangent angle is $180^\circ-60^\circ=120^\circ$, not 60°.
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  7. Problem 7
    Tangent Length·★★★☆☆
    Assess this claim as True or False: no matter where an outside point is chosen, its tangent length exceeds the radius of the circle.
    ▶Answer
    False
    ▶Step-by-step solution
    1. The tangent length depends on how far the external point is from the circle.
    2. A valid right-triangle configuration can have radius 12 units and tangent length 5 units.
    3. This counterexample has tangent length less than the radius, so the word ‘always’ makes the claim false.
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  8. Problem 8
    Tangent Length·★★★☆☆
    Assess this claim as True or False: when PT touches a circle centred at O, the segment PT is shorter than the centre-distance OP.
    ▶Answer
    True
    ▶Step-by-step solution
    1. Radius OT is perpendicular to tangent PT.
    2. Therefore triangle OPT is right-angled at T, making OP its hypotenuse.
    3. The hypotenuse is longer than either leg, so $PT<OP$.
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  9. Problem 9
    Angle Between Tangents·★★★★☆
    Assess this claim as True or False: a finite point outside a circle can have two tangent rays whose included angle is 0°.
    ▶Answer
    False
    ▶Step-by-step solution
    1. The two tangents touch the circle at distinct points.
    2. They are distinct intersecting lines through the external point.
    3. Their smaller angle is positive; 0° would mean that the two tangent rays coincide.
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  10. Problem 10
    Angle Between Tangents·★★★★☆
    Assess this claim as True or False: tangents from P enclose 90° around a circle of radius 9 cm, which makes P exactly $9\sqrt{2}$ cm from the centre.
    ▶Answer
    True
    ▶Step-by-step solution
    1. The centre line bisects the 90° angle between the equal tangents.
    2. One half is a 45°–45°–90° right triangle with the radius as a leg.
    3. Its hypotenuse is therefore $9\sqrt{2}$ cm, so the claim is true.
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  11. Problem 11
    Centres of Tangent Circles·★★★★★
    Assess this claim as True or False: fix A on segment PQ and allow circles tangent to PQ at A; every possible centre lies on the perpendicular bisector of PQ.
    ▶Answer
    False
    ▶Step-by-step solution
    1. The radius to A must be perpendicular to line PQ.
    2. Therefore the centres lie on the line through A perpendicular to PQ.
    3. That line is the perpendicular bisector of PQ only when A is the midpoint, so the general claim is false.
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  12. Problem 12
    Perpendicular Bisector of a Chord·★★★★★
    Assess this claim as True or False: as a circle varies while continuing to pass through fixed points P and Q, its centre remains on the perpendicular bisector of PQ.
    ▶Answer
    True
    ▶Step-by-step solution
    1. P and Q are points on each circle, so OP and OQ are radii of that circle.
    2. Hence every possible centre O is equidistant from P and Q.
    3. The locus of points equidistant from P and Q is their perpendicular bisector.
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  13. Problem 13
    Cyclic Quadrilaterals·★★★★☆
    A circle has centre O. From an outside point P, the contact points of the two tangents are Q and R. Show that the four points Q, O, R, and P lie on one circle.
    ▶Answer
    Proved
    ▶Step-by-step solution
    1. A radius is perpendicular to the tangent at its point of contact, so $\angle OQP=90^\circ$.
    2. Similarly, $\angle ORP=90^\circ$.
    3. These opposite angles sum to 180°, so QORP is a cyclic quadrilateral.
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  14. Problem 14
    Equal Tangents·★★★★☆
    An outside point B is joined tangentially to a circle at C and D, with each tangent measuring 7 cm. The angle between them is $120^\circ$. Establish that the distance from B to the centre O is twice BC.
    ▶Answer
    Proved
    ▶Step-by-step solution
    1. Triangles OCB and ODB are congruent, so BO bisects the 120° angle.
    2. Thus $\angle CBO=60^\circ$, while OC is perpendicular to BC.
    3. In right triangle BCO, $\cos60^\circ=BC/BO=1/2$, hence $BO=2BC$.
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  15. Problem 15
    Centres of Tangent Circles·★★★★☆
    Two non-parallel lines are both tangent to the same circle. Show that the circle's centre must be on one of the bisectors of the angle formed by the lines.
    ▶Answer
    Proved
    ▶Step-by-step solution
    1. Draw radii from the centre O to the two points of contact.
    2. Both radii are perpendicular to their respective lines and have equal length.
    3. Therefore O is equidistant from the two lines; the locus of such points is an angle bisector.
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  16. Problem 16
    Common Tangents·★★★★★
    From P, two rays are direct common tangents to a pair of unequal circles. Along the rays, the contact points occur in the orders P-A-B and P-C-D. Demonstrate that the portions between the circles, AB and CD, have the same length.
    ▶Answer
    Proved
    ▶Step-by-step solution
    1. Tangents from P to the first circle are equal, so PA = PC.
    2. Tangents from P to the second circle are equal, so PB = PD.
    3. Since AB = PA - PB and CD = PC - PD, subtraction gives AB = CD.
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  17. Problem 17
    Common Tangents·★★★★★
    For two congruent circles, mark the contact-to-contact portions of their direct common tangents as AB and CD. Show that these two portions are equal.
    ▶Answer
    Proved
    ▶Step-by-step solution
    1. The radii to A and B are equal and perpendicular to their common tangent.
    2. Together with the line of centres, they form a rectangle, so AB equals the distance between the centres.
    3. The same argument gives CD equal to that centre distance; hence AB = CD.
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  18. Problem 18
    Common Tangents·★★★★★
    Two common tangents cross at E between a pair of circles. Their complete contact-to-contact segments are AB and CD. Use tangent lengths from E to show that these segments are equal.
    ▶Answer
    Proved
    ▶Step-by-step solution
    1. From E, the tangent segments to the first circle are equal: EA = EC.
    2. The tangent segments to the second circle are also equal: EB = ED.
    3. Since AB = AE + EB and CD = CE + ED, the two common-tangent segments are equal.
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