Chapter 10: Circles

Key Theorems:
• Tangent is perpendicular to radius at point of contact
• Lengths of tangents from an external point are equal
• Two tangents can be drawn from an external point
• Angle between tangents from P = 180° - (angle subtended at centre)

Exercise 10.1

Q1. How many tangents can be drawn through a point inside a circle?

Solution

Zero. Tangents can only be drawn from a point on or outside the circle.

Q2. How many tangents through a point on the circle?

Solution

Exactly one. The tangent at any point on a circle is unique.

Q3. How many tangents through a point outside the circle?

Solution

Exactly two. Two tangents can be drawn from an external point.

Exercise 10.2

Q1. From Q, tangent length=24cm, distance from centre=25cm. Find radius.

Solution

r²+24²=25²
r²+576=625
r²=49
r=7 cm

Q2. TP and TQ are tangents, ∠PTQ=60°. Find ∠OPQ.

Solution

TP=TQ → ΔTPQ is isosceles
∠TPQ=∠TQP=(180-60)/2=60°
OP ⊥ TP → ∠OPT=90°
∠OPQ=90°-60°=30°

Q3. PA and PB are tangents inclined at 80°. Find ∠POA.

Solution

In quadrilateral OAPB:
∠OAP=∠OBP=90°
∠AOB=360-90-90-80=100°
∠POA=100/2=50°

Q4. Prove tangent segments from external point are equal.

Solution

Let PA and PB be tangents from P.
In ΔOAP and ΔOBP:
OA=OB (radii)
OP=OP (common)
∠OAP=∠OBP=90°
By RHS congruence, ΔOAP ≅ ΔOBP
∴ PA=PB

Q6. Prove ∠PTQ=2∠OPQ.

Solution

Let ∠OPQ=x
In ΔOPQ: OP=OQ → ∠OQP=x
In ΔTPQ: ∠TPQ=∠TQP=90°-x
∠PTQ=180-2(90-x)=2x
∴ ∠PTQ=2∠OPQ

Q7. Tangents at ends of a diameter are parallel. Prove.

Solution

Let AB be diameter. Tangents at A and B are perpendicular to AB. Since both are perpendicular to the same line AB, they are parallel.
∴ Tangents at ends of diameter are parallel.

Q8. Angle between tangents is supplementary to angle subtended at centre.

Solution

In quadrilateral OAPB:
∠OAP=∠OBP=90°
∠AOB+∠APB+90+90=360
∠AOB+∠APB=180
∴ Supplementary.

Q11. Tangent PQ at P, radius 5cm, OQ=12cm. Find PQ.

Solution

OP=5cm, OQ=12cm
PQ²=OQ²-OP²=144-25=119
PQ=√119 cm

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