CIRCLES

EXERCISE 10.1

Q1. How many tangents can a circle have?

Answer:
A circle can have infinitely many tangents.

• From a point outside the circle: 2 tangents can be drawn.

• From a point on the circle: 1 tangent can be drawn.

• From a point inside the circle: No tangent can be drawn.

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Q2: Fill in the blanks :

(i) A tangent to a circle intersects it in one point.
(ii) A line intersecting a circle in two points is called a secant.
(iii) A circle can have two parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called the point of contact.

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Q3: 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 (simple):
OP ⟂ PQ, so triangle OPQ is right-angled at P.

PQ = √(OQ² – OP²)
= √(12² – 5²)
= √(144 – 25)
= √119 cm

Answer: √119 cm (D)

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Q4: Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Answer (Construction steps):

1. Draw a circle with any centre O.

2. Draw a line L (the given line).

3. Draw a line parallel to L such that it touches the circle at exactly one point → this is the tangent.

4. Draw another line parallel to L such that it cuts the circle at two points → this is the secant.

The tangent line touches the circle at only one point, while the secant line passes through the circle intersecting it at two points.

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EXERCISE 10.2

Q1: 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:
Radius to the point of contact is perpendicular to the tangent, so triangle OPQ is right-angled at P.

r² + (tangent)² = (OQ)²
r² + 24² = 25²
r² = 625 − 576 = 49
r = 7 cm

Answer: 7 cm (A)

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Q2: In Fig. 10.11 (refer book), if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to ______.

Solution:
Join OP and OQ. Radii OP and OQ are perpendicular to tangents TP and TQ respectively, so PT ⟂ OP and QT ⟂ OQ.
Hence the angle between the tangents PT and QT = 180° − ∠POQ.
So ∠PTQ = 180° − 110° = 70°.

Answer: 70°. (B)

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Q3: If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to ______.

Solution:

• Angle between tangents (∠APB) = 80°.

• The angle between the lines joining centre O to points of contact (∠AOB) = 180° − ∠APB
  = 180° − 80°
  = 100°.

• In isosceles triangle OAP, OA = OP (radii and tangent–radius relation).
  So ∠POA = ½ ∠AOB
  = ½ × 100°
  = 50°.

Answer: 50°. (A)

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Q4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Proof:

1. Let AB be the diameter of a circle with centre O.

2. Draw tangents at the ends of the diameter, i.e., at points A and B.

3. Radius OA is perpendicular to the tangent at A, and radius OB is perpendicular to the tangent at B.
⇒ Tangent at A ⟂ OA and Tangent at B ⟂ OB.

4. Since OA and OB lie on the same straight line (diameter AB), we have:
OA ∥ OB (collinear line).

5. Therefore, two lines perpendicular to the same straight line are parallel.

Hence, the tangents drawn at the ends of a diameter of a circle are parallel.

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Q5: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Proof:

1. Let a circle have centre O and let a line ℓ be tangent to the circle at point P.

2. Suppose from O we drop a perpendicular to the tangent ℓ and let the foot be Q.

3. If Q = P then OQ is perpendicular to ℓ and passes through the centre — done.

4. Assume, for contradiction, that Q ≠ P. Then OQ < OP (because Q is the perpendicular projection of O on the line ℓ, so it is the nearest point on ℓ to O).

5. But OP is the radius of the circle. So OQ < OP means the distance from O to Q is less than the radius — hence Q lies inside the circle.

6. If Q is inside the circle and Q lies on line ℓ, then line ℓ must meet the circle at more than one point, contradicting the fact that ℓ is a tangent (a tangent meets the circle in exactly one point).

7. This contradiction shows Q cannot be different from P. Therefore Q = P, i.e. the line through O and P is perpendicular to the tangent at P.

Conclusion: The perpendicular at the point of contact to the tangent passes through the centre O.

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Q6: The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution:
Let O be the centre of the circle and P be the point of contact.
We know OP ⟂ AP.
So, in right triangle OAP:

OA2=OP2+AP2OA^2 = OP^2 + AP^2

5² = r² + 4²
25 = r² + 16
r² = 9
r = 3 cm

Answer: 3 cm

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Q7: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution:

1. Let O be the common centre of the two circles.
Radius of larger circle = 5 cm, radius of smaller circle = 3 cm.

2. Let AB be the chord of the larger circle that touches the smaller circle at point P.

3. Since AB touches the smaller circle, OP ⟂ AB and OP = 3 cm.

4. In right triangle OAP,
OA2=OP2+AP2OA^2 = OP^2 + AP^2.
Here OA = 5 cm, OP = 3 cm.
5² = 3² + AP²
25 = 9 + AP²
AP² = 16
AP = 4 cm.

5. Therefore, chord length AB = 2 × AP = 2 × 4 = 8 cm.

Answer: 8 cm

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Q8: A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12 (refer book). Prove that AB + CD = AD + BC

Proof:
Let the circle touch the sides of quadrilateral ABCD as follows:

• Side AB at P

• Side BC at Q

• Side CD at R

• Side DA at S

From the property of tangents drawn from an external point:

• AP = AS

• BP = BQ

• CQ = CR

• DR = DS

Now, write the sides in terms of these tangent lengths:

• AB = AP + PB

• BC = BQ + QC

• CD = CR + RD

• DA = DS + SA

So,
AB + CD = (AP + PB) + (CR + RD)
= (AP + CR) + (PB + RD)

AD + BC = (DS + SA) + (BQ + QC)
= (DS + BQ) + (SA + QC)

But AP = AS, BP = BQ, CQ = CR, DR = DS.
Therefore, AB + CD = AD + BC.

Hence proved.

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Q9: In Fig. 10.13 (refer book), XY and X′Y′ are two parallel tangents to a circle with centre O. Another tangent AB touches the circle at C and meets XY at A and X′Y′ at B. Prove that ∠AOB = 90°.

Solution:

1. XY and X′Y′ are parallel tangents, so the line AB intersects them at A and B.

2. Join OA and OB.

3. The tangent AB touches the circle at C. So, OC is perpendicular to AB.

4. The line OC passes through the centre O and meets AB at right angles.

5. This makes triangle OAC and OBC symmetric about OC.

6. Hence, OA and OB are perpendicular to each other.

Therefore, ∠AOB = 90°.

Answer: ∠AOB = 90°

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Q10: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Solution:

Given:
A circle with centre O. From an external point P, two tangents PA and PB touch the circle at A and B respectively.

To prove:
∠APB (the angle between the tangents) is supplementary to ∠AOB (the angle subtended by chord AB at the centre).
That is, ∠APB + ∠AOB = 180°.

Facts used:

1. A radius drawn to the point of contact is perpendicular to the tangent.
Therefore, OA ⟂ PA and OB ⟂ PB.
So, ∠OAP = 90° and ∠OBP = 90°.

Proof:

1. Consider quadrilateral AOPB.
Its interior angles are: ∠OAP = 90°, ∠OBP = 90°, ∠AOB, and ∠APB.

2. Sum of interior angles of a quadrilateral is 360°.
Therefore: ∠AOB + ∠APB + 90° + 90° = 360°.

3. Simplifying: ∠AOB + ∠APB + 180° = 360°.
Hence: ∠AOB + ∠APB = 180°.

Conclusion:
∠APB is supplementary to ∠AOB.
This proves that the angle between the two tangents from an external point is supplementary to the angle subtended at the centre by the line segment joining the points of contact.

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Q11: Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

Given a parallelogram ABCD that circumscribes a circle. Let the circle touch AB, BC, CD and DA at P, Q, R and S respectively.

Put lengths: AP = x, PB = y, BQ = y, QC = z, CR = z, RD = w, DS = w, SA = x.
(These equalities follow because tangent segments from the same vertex are equal.)

Now express the sides:
AB = AP + PB = x + y,
BC = BQ + QC = y + z,
CD = CR + RD = z + w,
DA = DS + SA = w + x.

Because ABCD is a parallelogram, opposite sides are equal:
AB = CD so x + y = z + w. (1)
BC = DA so y + z = w + x. (2)

Subtract (2) from (1):
(x + y) − (y + z) = (z + w) − (w + x)
This simplifies to x − z = z − x, so 2(x − z) = 0. Hence x = z.

Substitute x = z into (1): x + y = x + w, so y = w.

Therefore the four side-lengths become:
AB = x + y,
BC = y + x = x + y,
CD = x + y,
DA = y + x = x + y.

All four sides are equal. Hence ABCD is a rhombus. Hence proved

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Q12: A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14 (refer book)). Find the sides AB and AC.

Solution:

Let BC = a, AC = b, AB = c and let s be the semiperimeter of triangle ABC.

Given: BD = 8 cm and DC = 6 cm, so a = BC = BD + DC = 8 + 6 = 14 cm.
Also given inradius r = 4 cm.

For a triangle with an incircle, the lengths from the vertices to the points of contact satisfy
s − b = BD = 8 and s − c = DC = 6.
Hence b = s − 8 and c = s − 6.

Area of the triangle is Δ = r × s = 4s.

By Heron’s formula,
Δ = √[ s (s − a) (s − b) (s − c) ].
Here s − b = 8 and s − c = 6 and a = 14, so
Δ = √[ s (s − 14) × 8 × 6 ] = √[ 48 s (s − 14) ].

Equate the two expressions for Δ:
(4s)² = 48 s (s − 14)
16 s² = 48 s (s − 14)

Divide both sides by s (s > 0):
16 s = 48 (s − 14)
16 s = 48 s − 672
672 = 32 s
s = 21.

Now find b and c:
b = s − 8 = 21 − 8 = 13 cm,
c = s − 6 = 21 − 6 = 15 cm.

Hence AC = 13 cm and AB = 15 cm.

Check: s − a = 21 − 14 = 7, so
Δ = √[21 × 7 × 8 × 6] = √[7056] = 84.
Then r = Δ / s = 84 / 21 = 4 cm, which matches the given inradius.

Answer: AB = 15 cm and AC = 13 cm.
Hence proved.

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Q13: Prove that opposite interior angles of a quadrilateral that circumscribes a circle are supplementary.

Let ABCD be a quadrilateral which circumscribes a circle with centre O. Let the circle touch AB, BC, CD and DA at P, Q, R and S respectively.

1. Radii to the points of contact are perpendicular to the sides.
So OP is perpendicular to AB, OQ to BC, OR to CD and OS to DA.

2. The angle at a vertex equals the angle between the perpendiculars at the two adjacent contact points.
Thus angle A equals angle between OP and OS, so angle A = ∠POS.
Similarly, angle B = ∠POQ, angle C = ∠QOR and angle D = ∠ROS.

3. All four central angles about O add to 360 degrees:
∠POQ + ∠QOR + ∠ROS + ∠SOP = 360°.

4. Group these four angles into two pairs of opposite central angles:
(∠SOP + ∠QOR) + (∠POQ + ∠ROS) = 360°.

But ∠SOP = angle A and ∠QOR = angle C (from step 2).
Also ∠POQ = angle B and ∠ROS = angle D.

So (angle A + angle C) + (angle B + angle D) = 360°.

5. A quadrilateral has four interior angles summing to 360°, so angle A + angle B + angle C + angle D = 360°.
Comparing with the equality from step 4 we see that angle A + angle C and angle B + angle D must together make 360°. By symmetry of the two pairings, each pair must sum to 180°. In particular,
angle A + angle C = 180°.

Therefore opposite interior angles A and C are supplementary. By the same argument B + D = 180°.

Hence proved.

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