Circles Exercise 12.1 |

**Q. 1: Find the length of the tangent drawn to a circle with radius 8cm from a point 17cm away from the center of the circle.**

**Sol:**

PA is the tangent to the circle with center O and radius AO = 8 cm. The point P is at a distance of 17 cm from O.

In \(\Delta\) PAO, A = 90^{o}

By Pythagoras theorem:

PO^{2 }= PA^{2} + AO^{2}

Or PA^{2} = PO^{2} – AO^{2}

=> PA = \(\sqrt{(17)^{2}-(8)^{2}}\) cm

= \(\sqrt{289-64}\) cm

= \(\sqrt{225}\) cm = 15 cm

Hence, the length of the tangent is 15 cm.

**Q. 2: A point P is 25cm away from the center of a circle and the length of tangent drawn from P to the circle is 24cm. Find the radius of the circle.**

**Sol: **

PA is the tangent to the circle with center O and radius, such that PO = 25 cm, PA = 24cm

In PAO, A = 90^{o}

By Pythagoras theorem:

PO^{2} = PA^{2} + AO^{2}

Or OA^{2} = PO^{2} – PA^{2}

=> OA = (25)^{ 2} – (24)^{ 2} cm

= (25 + 24) (25-24) cm =49 cm

Therefore, OA = 7 cm.

Hence, the radius of the circle is 7 cm.

**Q. 3: Two concentric circles are the radii 6.5cm and 2.5cm. Find the length of the chord of the larger circle which touches the smaller circle.**

**Sol: **

Given AP is a tangent at A and OA is radius through A and PA and PB are the tangent segments to circle with center O.

Therefore, OA is perpendicular to AP, similarly, OB is perpendicular to BP.

Therefore, \(\angle\) OAP = 90^{o}

And \(\angle\) OBP = 90^{o}

So, \(\angle\) OAP = \(\angle\) OBP = 90^{o}

Therefore, \(\angle\)OBP + \(\angle\)OAP = (90^{o }+ 90^{o}) = 180^{o}

Thus, the sum of opposite angles of quad. AOBP is 180^{o}

Hence, AOBP is a cyclic quadrilateral.

**Q. 4: In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12cm, BC = 8cm and AC = 10cm, find the lengths of AD, BE and CF.**

**Sol:**

Given: From an external point P, tangent PA and PB are drawn to a circle with center O. CD is the tangent to the circle at a point E and PA = 14 cm.

Since the tangents from an external point are equal, we have PA = PB,

Also, CA = CE and DB = DE

Perimeter of \(\Delta\) PCD = PC + CD + PD

= (PA – CA) + (CE + DE) + (PB – DB)

= (PA – CE) + (CE + DE) + (PB – DE)

= (PA + PB) = 2PA = (2 x 14) cm = 28 cm.

Hence, Perimeter of \(\Delta\) PCD = 28 cm.

**Q. 5: In the given figure, PA and PB are the tangents to a circle with center O. Show that the points A, O, B and P are concyclic.**

**Sol:**

A circle is inscribed in a triangle ABC touching AB, BC and CA at P, Q and R respectively.

Also, AB = 10 cm, AR = 7 cm, CR = 5 cm

AR and AP are the tangents to the circle.

AP = AR = 7cm

AB = 10 cm

BP = AB – AP = (10 – 7) = 3 cm

Also, BP and BQ are tangents to the circle.

BP = BQ = 3 cm

Further, CQ and CR are tangents to the circle.

CQ = CR = 5 cm.

BC = BQ + CQ = (3 + 5) cm = 8 cm.

Hence, BC = 8 cm

**Q. 6: In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.**

**Sol:**

Let the circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively.

We know that the lengths of tangents drawn from an exterior point to a circle are equal.

AP = AS – – – – – – – – – – (1) {tangents from A}

BP = BQ – – – – – – – – – – (2) {tangents from B}

CR = CQ – – – – – – – – – – – (3) {tangents from C}

DR = DS – – – – – – – – – – – -(4) {tangents from D}

Adding (1), (2) and (3) we get:

AP + BP + CR + DR = AS + BQ + CQ + DS

=> (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

=> AB + CD = AD + BC

=> AD = (AB + CD) – BC = {(6+4) – 7} cm = 3 cm

Hence, AD = 3 cm.

**Q.7: From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14cm, find the perimeter of ⧍PCD.**

**Sol:**

**Given:** Two tangents segments BC and BD are drawn to a circle with center O such that \(\angle\)CBD = 120^{o}

Join OB, OC and OD.

In triangle OBC,

\(\angle\) OBC = \(\angle\) OBD = 60^{o}

\(\angle\) = 90^{o} (BC is tangent to the circle)

Therefore, \(\angle\) BOC = 30^{o}

\(\frac{BC}{OB}\) = sin 30^{o} = \(\frac{1}{2}\)

=> OB = 2BC

**Q. 8: A circle is inscribed in a ⧍ABC touching AB, BC and AC at P, Q and R respectively. If AB = 10cm, AR = 7cm and CR = 5cm, find the length of BC.**

**Sol: **

Given O is the center of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. PA = 10 cm. Join OA, OB and OP.

Then, OB = 4 cm, OA = 6 cm, PA = 10 cm.

In triangle OAP,

OP^{2} = OA^{2} + PA^{2} = (6)^{2} + (10)^{2} = 136 cm^{2}

In \(\Delta\) OBP

BP = \(\sqrt{OP^{2}-OB^{2}}=\sqrt{136-16}\) cm = \(\sqrt{120}\) cm = 10.9 cm.

Hence, BP = 10.9 cm

**Q. 9: In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6cm, BC = 7cm and CD = 4cm. Find AD.**

**Sol:**

Join OR and OS, then OR = OS

OR \(\perp\) DR and OS \(\perp\) DS

Therefore, ORDS is a square.

Tangents from an external point being equal, we have

BP = BQ

CQ = CR

DR = DS

Therefore, BQ = BP = 27 cm

=> BC – CQ = 27 cm

=> 38 – CQ = 27

=> CQ = 11 cm

=> CR = 11 cm

=> CD – DR = 11 cm

=> 25 – DR = 11 cm

=> DR = 14 cm

=> r = 14 cm

Hence, radius = 14 cm

**Q.10:** **In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a a circle. Prove that the point of contact P bisects the base BC**

**Sol: **

Here, OA=OB and OA is perpendicular to AP,

OA is perpendicular to BP (Since tangents drawn from an external point are perpendicular to the radius at the point of contact)

∠OAP = 900, ∠OBP = 900

∠OAP+∠OBP=900+900=1 800

∠A0B+∠APB=1 800 (Since, ∠OAP+∠OBP+∠AOB+∠APB = 3600)

Sum of opposite angle of a quadrilateral is 180°.

Hence, A, O, B and P are concyclic.

**Q.11: In the given figure, O is the centre of two concentric circles of radii 4cm and 6cm respectively. PA and PB are tangents to the outer and inner circle respectively. If PA = 10cm, find the length of PB up to one place of decimal.**

**Sol:**

We know that tangent segments to a circle from the same external point are congruent.

Now, we have

AR = AQ, BR = BP and CP = CQ

Now, AB = AC

=> AR + RB = AQ + QC

=> AR + RB = AR + QC

=> RB = QC

=> BP = CP

Hence, P bisects BC at P.

**Q.12: In the given figure, O is the centre of two concentric circles of radii 4cm and 6cm respectively. PA and PB are tangents to the outer and inner circle respectively. If PA = 10cm, find the length of PB up to one place of decimal.**

**Sol:**

**Construction:** Join OA, OB, OC, OE perpendicular to AB at E and OF perpendicular to AC at F

We know that tangent segments to a circle from the same external point are congruent.

Now, we have,

AE = AF, BD = BE = 6 cm and CD = CF = 9 cm

Now,

Area ⧍ABC=Area ⧍BOC + Area ⧍AOB + Area ⧍AOC =>54

= 12 x BC x OD + 12xABx0E + 12 x AC x OF =>108

=>36=30+2x => x=3 cm

AB = 6 + 3 = 9cm

AC = 9 +3 = 12 cm.

**Q.13: PQ is a chord of length 4.8cm of a circle of radius 3cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.**

**Sol:**

Let TR = y and TP = x

We know that the perpendicular drawn from the centre to the chord bisects it.

PR = RQ

Now, PR + RQ = 4.8

=>PR + PR = 4.8

=> PR = 2.4

Now, in right triangle POR:

By Using Pythagoras theorem, we have

PO^{2} = OR^{2} + PR^{2}

=> 3^{2} = OR^{2} + (2.4)^{2}

OR^{2} = 3.24

OR = 1.8

Now, in right triangle TPR

By Using Pythagoras theorem, we have:

TP^{2} = TR^{2 }+ PR^{2} + x^{2} = y^{2} + (2.4)^{ 2}

x^{2} = y^{2} + 5.76 ……..(1)

Again, in right triangle TPQ

By Using Pythagoras theorem, we have:

TO^{2} = TP^{2} + PO^{2}

(y + 1.8)^{ 2} = x^{2} + 3^{2}

y^{2} + 3.6y + 3.24 = x^{2} + 9

y^{2} + 3.6y = x^{2} + 5.76 ……..(2)

Solving (1) and (2), we get:

x = 4 cm and y = 3.2 cm

TP = 4cm.

**Q.14: Prove that the line segment joining the points of contact of two parallel tangents of a circle passes through its centre.**

**Sol:**

Suppose CD and AB are two parallel tangents of a circle with centre O

**Construction: **

Draw a line parallel to CD passing through O i.e, OP

We know that the radius and tangent are perpendicular at their point of contact. ∠OQC = ∠ORA = 90°

Now, ∠OQC + ∠POQ = 180° (co-interior angles)

=> ∠POQ = 180° – 90° = 90°

Similarly, Now, ∠ORA + ∠POR = 180° (co-interior angles)

∠POR = 180° – 90° = 90°

Now, ∠POR + ∠POO = 90° + 90° = 180°

Since ∠POR and ∠POQ are linear pair angles whose sum is 180°

Hence, QR is a straight line passing through centre O.

**Q.15: In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R, S respectively. If AB = 29cm, AD = 23cm, ∠B = 90° and DS = 5cm then find the radius of the circle.**

**Sol: **We know that tangent segments to a circle from the same external point are congruent.

Now, we have

DS = DR, AR = AQ

Now, AD = 23 cm

=> AR + RD = 23

=> AR = 23 – RD

=> AR=23- 5

=> AR = 18 cm

Again, AB = 29 cm

=> AQ+Q6=29

=> QB=29 -AQ

=> QB = 29 – 18

=> QB = 11 cm

Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ. Hence, BQOP is a square.

We know that all the sides of the square are equal.

Therefore, BQ = PO = 11cm

Hence, the radius of the circle is 11 cm.

Related Links |
||
---|---|---|

NCERT Books | NCERT Solutions | RS Aggarwal |

Lakhmir Singh | RD Sharma Solutions | NCERT Solutions Class 6 to 12 |