OP Malhotra Class 10 Maths Solutions Chapter 14 Circle Ex 14(e)

Students can cross-reference their work with S Chand Class 10 Maths Solutions ICSE Chapter 14 Circle Ex 14(e) to ensure accuracy.

S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(e)

Question 1.
In fig., if AB and CD are two chords of a circle intersecting at a point P inside the circle such that
(i) AP = 8 cm, CP = 6 cm and PD = 4 cm, find PB.
(ii) AB = 12 cm, AP = 2 cm and PD = 4 cm, find CP.
(iii) AP = 6 cm, PB = 5 cm and CD = 13 cm, find CP.

Solution:
∵ AB and CD are two chords which intersect
at P inside the circle
∴ AP × PB = CP × PD

(i) Now AP = 8 cm, CP = 6 cm and PD = 4 cm
∴ 8 × PB = 6 × 4 ⇒ PB = \(\frac{6 \times 4}{8}\) cm
∴ PB = 3 cm

(ii) AB = 12 cm, AP = 2 cm and PD = 4 cm
∴ PB = AB – AP = 12 – 2 = 10 cm
Now AP × PB = CP × PD
⇒ 2 × 10 = CP × 4 ⇒ CP = \(\frac{2 \times 10}{4}\) = 5 cm
∴ CP = 5 cm

(iii) AP = 6 cm, PB = 5 cm and CD = 13 cm
Let CP = x, then PD = CD – CP
= (13 – x) cm
Now AP × PB = CP × PD
⇒ 6 × 5 = x(13 – x) ⇒ 30 = 13 x – x²
⇒ x² – 13x + 30 = 0
⇒ x² – 10x – 3x + 30 = 0

⇒ x(x – 10) – 3 (x – 10) = 0
⇒ (x- 10)(x – 3) = 0
Either x – 10 = 0, then x = 10
or x – 3 = 0, then x = 3
CP = 10 cm or 3 cm

Question 2.
In figure, if AB and CD are two chords of a circle which when produced meet at a point P such that
(i) PA = 10 cm, PB = 4 cm and PC = 8 cm, find PD.
(ii) PC = 15 cm, CD = 7 cm and PA = 12 cm, find AB.
(iii) PA = 16 cm, PC = 10 cm and PD = 8 cm, find AB.

Solution:
∵ Chords AB and CD of a circle intersect each other at P outside the circle
∴ AP × PB = CP × PD

(i) PA = 10 cm, PB = 4 cm and PC = 8 cm
Now AP × PB = CP × PD
⇒ 10 × 4 = 8 × PD
⇒ PD= \(\frac{10 \times 4}{8}\) = 5
∴ PD = 5 cm

(ii) PC = 15 cm, CD = 7 cm and PA = 12 cm
PD = CP – CD = 15 – 7 = 8 cm
Now PA × PB = PC × PD
12 × PB = 15 × 8 ⇒ PB = \(\frac{15 \times 8}{12}\) = 10 cm
∴ AB = PA – PB = 12 – 10 = 2 cm

(iii) PA = 16 cm, PC = 10 cm and PD = 8 cm
Now PA × PB = PC × PD
⇒ 16 × PB = 10 × 8
⇒ PB = \(\frac{10 \times 8}{16}\) = 5
∴ AB = PA – PB = 16 – 5 = 11 cm

Question 3.
(i) In figure, if PT is a tangent to the circle, PB = 4 cm and AB = 12 cm, then PT = …….cm.

(ii) From an external point P, the tangent PT and a secant PAB are drawn. If PA = 9.6 cm and PB = 2.4 cm, determine PT.
Solution:
(i) In the figure, PT is tangent and PBA is a secant to the circle PB = 4 cm, AB = 12 cm

∴ PA = PB + AB = 4 + 12 = 16 cm
∴ PT² = PA × PB
= 16 × 4 = 64 = (8)²
∴ PT = 8 cm

(ii) From an external point P, PT is the tangent to the circle and PAB is a secant

∴ and PA = 9.6 cm, PB = 2.4 cm
Now PT² = PA × PB = 9.6 × 2.4
= 23.04 = (4.8)²
∴ PT = 4.8 cm

Question 4.
The angle A of the triangle ABC is a right angle. The circle on AC as diameter cuts BC at D. If BD = 9, and DC = 7, calculate the length of AB.

Solution:
In △ABC, ∠A = 90°
A circle is drawn on AC as diameter which intersects BC at D
AD is joined
BD = 9 and DC = 7

We see that in the circle AB is the tangent at A and BDC is the secant of the circle
∴ AB² = BD × BC
= 9 × 16 {∵ BC = BD + DC = 9 + 7 = 16}
= 144 = (12)²
∴ AB = 12

Question 5.
In △ABC, AB = 9, AC = 12, F is the mid-point of AC; the circle BFC intersects AB at E; find BE.
Solution:
Steps of construction :
(i) Draw a line segment AB = 9
At A draw ∠BAC = 90° and cut off AC = 12
(ii) Join BC. Take F as midpoint of AC.
(iii) Now draw perpendicular bisector of FC and BC intersecting each other at O.
(iv) With centre O, and radius OC draw a circle with passes through B, F and C and intersects AB at E on measuring BE = 1.

Question 6.
In △ABC, ∠BAC = 90°, AB = 4, AC = 3, AD is an altitude, find BD.
Solution:
ABC ABC is a right angled triangle in which ∠A = 90°,
AD ⊥ BC
AB = 4, AC = 3

∴ BC² = AB2+AC2 (Pythagoras theorem)
= (4)² + (3)² = 16 + 9 = 25 = (5)²
∴ BC = 5
Let BD = x, then DC = 5 – x
Now AB² = BD × BC
(4)² = x × 5 ⇒16 = 5x ⇒ x = \(\frac { 16 }{ 5 }\) = 3\(\frac { 1 }{ 5 }\)
∴ BD = 3\(\frac { 1 }{ 5 }\)

Question 7.
From the external point P, PA is a tangent to the circle at A. PBC is a secant intersecting the circle at B and C. What is the power of P with respect to the circle if PA = 7 ? What is the value of PB.PC ?
Solution:
From a point P,
Outside the circle PA is tangent and PBC is the secant PA = 7

Power of P with respect to the circle
= PA² = (7)² = 49
∵ PA is tangent and PBC is the secant
∴ PA² = PB × PC => 49 = PB × PC
Hence PB.PC = 49

Question 8.
In figure, ABC is a triangle inscribed in a circle. AB =AC = 10 cm, BC = 16 cm. The chord AE is at right angles to the chord BC at D. Calculate DE and the radius of the circle.

Solution:
△ABC in which AB = AC = 10 cm and BC = 16 cm, is inscribed in a circle AE is a chord which is at right angle to BC at D
In △ABC,
AB = AC and AD ⊥ BC
∴ BD = DC = \(\frac { 16 }{ 2 }\) = 8 cm
Now in right △ABD,
AB² = BD² + AD² ⇒ (10)² = (8)² + AD²
⇒ 100 = 64 + AD²
⇒ AD² = 100 – 64 = 36 = (6)²
∴ AD = 6
Now two chords AE and BC intersect each other at D
BD × DC = AD × DE
⇒ 8 × 8 = 6 × DE ⇒ DE = \(\frac{8 \times 8}{6}\) = \(\frac{64}{6}\) = \(\frac{32}{3}\)
∴ DE = \(\frac{32}{3}\) = 10\(\frac{2}{3}\) cm
Now AE = AD + DE = 6 + 10\(\frac{2}{3}\) = 16\(\frac{2}{3}\) cm
∴Radius = \(\frac{1}{3}\) AE (∵ AE is diameter)
= \(\frac{1}{3}\) × 16\(\frac{2}{3}\) = 8\(\frac{1}{3}\) cm

Question 9.
In figure, XY is a tangent to the circle with centre O. XCD is a secant. Calculate r, the radius of the circle.

Solution:
In the figure, XY is a tangent to the circle with centre O. XCD is a secant
ON ⊥ DX
Join OC and join OB
CX = 4 and XB = 6
∵XB is tangent and XCD is the secant
Let CD = x
∴ XB² = XC × XD ⇒ (6)² = 4 (4 + x)
= 4(4 + x) = 36 ⇒ 4 + x = \(\frac{36}{4}\) = 9
x = 9 – 4 = 5
∴ CD = 5
∵ ON ⊥ CD
∴ CN = ND = \(\frac{5}{2}\)
∴ Radius OB = NX = CN + XC
= \(\frac{5}{2}\) + 4 = 6\(\frac{1}{2}\) cm

Question 10.
In figure, O is the centre of the circle. If
(i) AX = 5 cm, XD = 7 cm, CX = 10 cm; find BX.
(ii) OA = 6 cm, BX = 5 cm, OX = 4 cm; find XC.
(iii) CD = 2 cm, DP = 6 cm, BP = 3 cm; find AB.

Solution:
In the figure, PBA, PDC are the secants PT is the tangents to the circle with centre O AD and BC are joined which intersect each other at X
(i) AX = 5 cm, XD = 7 cm, CX = 10 cm
∵ Two chords AD and BC intersect each other at X
∴ AX.XD = CX.XB
⇒ 5 × 7 = 10 × XB
⇒ XB = \(\frac{5 \times 7}{10}\) = \(\frac{7}{2}\) = 3.5 cm

(ii) OA}=6 cm, BX = 5 cm, OX = 4 cm
Join OA, OX and produce it to both sides meeting the circle at P and Q

OA = OP = OQ (radii)
XP = OP – OX = OA – OX = 6 – 4 = 2 cm
XQ = OX + OQ = OX + OA = 6 + 4 = 10 cm
∵ Chord BC and PQ intersect each other at X
∴ XC × 5 = 2 × 10 ⇒ XC = \(\frac{2 \times 10}{5}\) = 4
∴ XC = 4 cm

(iii) CD = 2 cm, DP = 6 cm, BP = 3 cm
Let AB = x
∵ Chords AB and CD intersect each other at P outside the circle

∴ PA × PB = PC × PD
(x + 3) × 3 = (2 + 6) × 6
(x + 3) × 3 = 8 × 6 ⇒ x + 3 = \(\frac{8 \times 6}{3}\) = 16
x = 16 – 3 = 13
∴ AB = 13 cm

Question 11.
In figure, two circles intersect each other at the points P and Q. If AB and AC are the tangents to the two circles from a point A on QP produced, show that AB = AC.

Solution:
Given : Two circles intersect each other at P and Q
AB and AC are the tangents on QP produced from A
To prove: AB = AC
Proof : ∵AB is the tangent and APQ is a secant
∴ AB² = AP × AQ …(i)
Similarly AC is the tangent and APQ is the secant
∴ AC² =AP × AQ …(ii)
From (i) and (ii)
AB² = AC²
⇒ AB = AC
Hence proved.

Question 12.
In figure, AB is any chord of a circle with centre O and P is any point on this chord. A perpendicular drawn through P on OP cuts the circumference in X. Prove that AP.PB = PX².

Solution:
Given : In a circle AB is a chord of a circle with centre O
P is any point on AB and PX ⊥ OP
To prove : AP.PB = PX²
Construction : Produce XP to meet the circle at Y

Proof : ∵ XY is a chord and OP ⊥ XY
∴ P is the midpoint of XY
∴ PX = PY
Now chord AB and XY intersect each other at P
∴ AP × PB = XP × PY
⇒ AP.PB = XP × XP (∵ XP = PY)
⇒ AP.PB = XP² or AP.PB = PX²
Hence proved.

Question 13.
In figure, the two circles intersects at S and T, and STP, BSC, BAP and CDP are st. lines. Prove that
(i) the quad. PATD is cyclic;
(ii) PA.PB = PD.PC.

Solution:
Given : Two circles intersect each other at T and S. STP, BSC, BAP and CDP are straight lines drawn
To prove :
(i) Quad. PATD is a cyclic
(ii) PA.PB = PD.PC
Construction : Join AT and TD
Proof:
(i) ∵ ABST is a cyclic quad.
∴ Ext. ∠ATP = Int. opp. ∠B (i)
Similarly in cyclic quad. CDTS,
Ext. ∠DTP = ∠C ….(ii)
Adding (i) and (ii)
∠ATP + ∠DTP = ∠B + ∠C ….(iii)
But in △PBC,
∠BPC + ∠B + ∠C= 180° (Sum of angles of a triangle)
⇒ ∠B + ∠C = 180° – ∠BPC From (iii)
∠ATP + ∠DTP =180°- ∠BPC
⇒ ∠ATD + ∠BPC = 180°
But these are sum of opposite angles of quad. PATD
∴ Quad. PATD is a cyclic quad.,

(ii) ∵ Chords BA and ST intersect at P outside the circle
∴ PA × PB = PT × PS ….(i)
Similarly CD and ST chords meet at P
∴ PD × PC = PT × PS ….(ii)
From (i) and (ii)
PA.PB = PD.PC Hence proved.

Question 14.
Two circles intersect at A. Chords PAQ and RAS are drawn through A, each passing through the centre of one of the two circles and terminated by the circum-ference. Prove PA.AQ = RA.AS.
Solution:
Given : Two circles intersect each other at A
Chords PAQ and RAS are drawn such that PAQ passes through O’ and RAS, passes through O, the centres of circles respectively
To prove : PA.AQ = RA.AS
Construction : Join PR and QS

Proof: In △APR and △ASQ,
∠P = ∠S (each 90° being in a semicircle)
∠PAR = ∠QAS (Vertically opposite angles)
∴ △APR ~ △ASQ (AA axiom)
∴ \(\frac{\text { PA }}{\text { AS }}\) = \(\frac{\text { RA }}{\text { AQ }}\)
∴ PA.AQ = RA.AS (By cross multiplication)
Hence proved.

Question 15.
Two circles intersect at points A and B. From a point P on the common chord BA produced, secants PCD and PEH are drawn one to each circle. Prove that the points C, D, H, E are concyclic.
Solution:
Given : Two circles intersect each other at A and B.
P is a point on BA produced through P, PCD and PEH are secant drawn to each circle.

To prove : C, D, H, E are concyclic
Proof : ∵ DC and BA are chords which intersect at P outside the first circle
∴ PC × PD = PA × PB ….(i)
Similarly, chords BA and HE intersect each other at P outside the second circle
∴ PA × PB = PE × PH ….(ii)
From (i) and (ii)
PC × PD = PE × PH
But there are two chords DC and HE which meet at P outside the circle
∴ C, D, H and E are concyclic Hence proved.

Question 16.
In figure, PB = BT and PT is tangent to the circle, then prove that
(i) △PTC is isosceles
(ii) PB.PC = TC²

Solution:
Given : In the figure, PB = BT PT is the tangent to the circle
PB is produced to meet the circle at C TC is joined
To prove :
(i) △PTC is an isosceles
(ii) PB.PC = TC²
Proof:
(i) In △PBT,
PB = BT (given)
∴ ∠1 = ∠2 ….(i)
But PT is tangent and BT is the chord of the circle
∴ ∠2 = ∠3 ….(ii)
(Angle in the alternate segment) From (0 and (ii)
∠1 = ∠3
∴ In ATPC,
TP = TC (Sides opposite to equal angles)
∴ △PTC is an isosceles triangle

(ii) ∵ PT is tangent and PBC is secant of the circle
∴ PT² = PB.PC
⇒ PB.PC = PT² = TC²
[∵ PT = TC proved in (i)]
⇒ PB.PC = TC²
Hence proved.