| 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 = 90o
By Pythagoras theorem:
PO2 = PA2 + AO2
Or PA2 = PO2 – AO2
=> 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 = 90o
By Pythagoras theorem:
PO2 = PA2 + AO2
Or OA2 = PO2 – PA2
=> 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 = 90o
And \(\angle\) OBP = 90o
So, \(\angle\) OAP = \(\angle\) OBP = 90o
Therefore, \(\angle\)OBP + \(\angle\)OAP = (90o + 90o) = 180o
Thus, the sum of opposite angles of quad. AOBP is 180o
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 = 120o
Join OB, OC and OD.
In triangle OBC,
\(\angle\) OBC = \(\angle\) OBD = 60o
\(\angle\) = 90o (BC is tangent to the circle)
Therefore, \(\angle\) BOC = 30o
\(\frac{BC}{OB}\) = sin 30o = \(\frac{1}{2}\)
=> 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,
OP2 = OA2 + PA2 = (6)2 + (10)2 = 136 cm2
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
PO2 = OR2 + PR2
=> 32 = OR2 + (2.4)2
OR2 = 3.24
OR = 1.8
Now, in right triangle TPR
By Using Pythagoras theorem, we have:
TP2 = TR2 + PR2 + x2 = y2 + (2.4) 2
x2 = y2 + 5.76 ……..(1)
Again, in right triangle TPQ
By Using Pythagoras theorem, we have:
TO2 = TP2 + PO2
(y + 1.8) 2 = x2 + 32
y2 + 3.6y + 3.24 = x2 + 9
y2 + 3.6y = x2 + 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.
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