Circles

Given that A and B are points on (O, r) and AB is a not a diameter of the circle.

We have to prove that the tangents to the circle at A and B are not parallel.

Proof:

Using the method of contradiction,

Let l and m be two parallel tangents to the circle have centre O drawn at the points A and B.

∴ OA ⊥ l and OB ⊥ m

Consider OA and OB perpendicular to l and m respectively and O is a common point.

∴ since l and m are two parallel lines, A – O – B

Hence, AB is a diameter, which contradicts with our assumption.

∴ Our assumption is wrong i.e. l and m are intersecting lines.

∴ Tangents to the circle at A and B are not parallel.

Hence proved.

Given that A and B are the points on (O, r) such that tangents at A and B intersect in P.

We have to prove that OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

Proof:

In circle (O, r), AP is a tangent at A and BP is the tangent at B.

⇒ ∠OAP = ∠OBP = 90°

Considering ΔOAP and ΔOBP,

⇒ OA = OB (radius)

⇒ OP = OP (common segment)

∴ By RHS theorem, ΔOAP = ΔOBP i.e. OAP and OBP is a congruence.

∴ ∠APO = ∠BOP and ∠AOP = ∠BOP

Here, O is in the interior part of ∠APB and P is in the interior part of ∠AOB.

∴ OP is the bisector of ∠AOB and PO is the bisector of ∠APB.

Hence proved.

Given that A and B are points on circle (O, r) such that tangents at A and B to the circle intersect in P.

We have to prove that the circle with OP as a diameter passes through A and B.

Proof:

We know that a tangent drawn to a circle is perpendicular to the radius drawn from the point of contact.

⇒ OA ⊥ AP and OB ⊥ PB

⇒ ∠OAP = 90° and ∠OBP = 90°

∴ ∠OAP + ∠OBP = 180° … (1)

For OAPB,

∠OAP + ∠APB + ∠AOB + ∠OBP = 360°

⇒ (∠APB + ∠AOB) + (∠OAP + ∠OBP) = 360°

From (1),

⇒ ∠APB + ∠AOB + 180° = 360°

∴ ∠APB + ∠AOB = 180° … (2)

From (1) and (2),

OAPB is cyclic.

Here, OP makes a right angle at A.

Then OP is a diameter.

∴ The circle with OP as a diameter passes through A and B.

Hence proved.

Given that circle (O, r₁) and circle (O, r₂) are such that r₁ > r₂.

∴ The circles are concentric.

Let chord AB of circle (O, r₁) touches circle (O, r₂) at P.

Thus, AB is tangent to circle (O, r₂).

∴ OP ⊥ AB and P ∈ AB

Here, P is the foot of the perpendicular drawn from centre O on the chord AB of circle (O, r₂).

∴ P is the midpoint of AB.

⇒ AB = 2AP … (1)

Consider right angle ΔOPA,

By Pythagoras Theorem,

⇒ OA² = AP² + OP²

⇒ r₁² = AP² + r₂²

⇒ AP² = r₁² – r₂²

∴ AP =

From (1),

∴ AB = 2

From example 4,

Given that r₁ = 41 and r₂ = 9

Length of chord AB = 2

⇒ AB = 2

= 2

= 2

= 2 (40)

= 80

∴ AB = 80

Given that P is the point in exterior of circle (O, r) and the tangents from P to the circle touch the circle at X and Y.

(1) Here OX is the radius of the circle and PX is the tangent.

∴ OX ⊥ PX

Given r = 12 = OX and XP = 5

For right angled ΔPXO,

⇒ OP² = PX² + OX²

= 5² + 12²

= 25 + 144

= 169

⇒ OP² = 169

∴ OP = 13

(2) Given ∠XOY = 110°

Here ∠XPY and ∠XOY are supplementary angles.

We know that two angles are supplementary when they add upto 180°.

⇒ ∠XPY + ∠XOY = 180°

⇒ ∠XPY + 110° = 180°

⇒ ∠XPY = 70°

Also, OP is a bisector pf ∠XPY

∴ ∠XPO = 1/2 ∠XPY = 1/2 (70°) = 35°

(3) OY is the radius of the circle and PY is a tangent.

∴ OY ⊥ PY

Now, OY = r,

Given that OP = 25 and PY = 24

In right angled ΔPYO,

By Pythagoras Theorem,

⇒ OP² = OY² + PY²

⇒ 25² = r² + 24²

⇒ r² = 625 – 576

⇒ r² = 49

∴ r = 7

(4) Given ∠XPO = 80°

Here, OX is the radius of the circle and PX is a tangent.

∴ OX ⊥ PX

Also, ∠XOP and ∠XPO are complementary angles.

We know that two angles are complementary when they add up to 90°.

⇒ ∠XOP + ∠XPO = 90°

⇒ ∠XOP + 80° = 90°

∴ ∠XOP = 10°

Given the radius of larger circle = 73 = OB and radius of smaller circle = 55 = OM

Since AB is a tangent, OM ⊥ AB.

Consider ΔOMB,

Here, ∠OMB is a right angle.

By Pythagoras Theorem,

⇒ OB² = OM² + MB²

⇒ MB² = OB² – OM²

= 73² – 55²

We know that a² – b² = (a + b) (a – b)

⇒ MB² = (73 + 55) (73 – 55)

= (128) (18)

= 2304

∴ MB = 48

Now, length of chord = AB = 2MB = 2 (48) = 96

∴ The length of chord is 96.

Given that AB is a diameter of circle (O, 10).

⇒ OA = OB = 10 = radius

⇒ AB = 20 = diameter

Also given a tangent is drawn from B to circle (O, 8) which touches the circle at D.

⇒ OD = 8 = radius

Since BD is a tangent, OD ⊥ BD.

And since the angle is inscribed in a semi circle, ∠ACB = 90°.

⇒ ∠ODB = ∠ACB = 90°

∴ ∠DBO ≅ ∠CBA

By AA theorem,

Thus, correspondence ODB ↔ ACB is a similarity.

Then =

⇒ =

⇒ AC = = 16

∴ AC = 16

Given OP = 34, PQ = 16

OQ is the radius of the circle.

Since PQ is a tangent to the circle, OQ ⊥ PQ.

In right angled ΔOQP,

By Pythagoras Theorem,

⇒ OP² = PQ² + OQ²

⇒ 34² = 16² + OQ²

⇒ OQ² = 1156 – 256

⇒ OQ² = 900

⇒ OQ = 30

∴ Diameter of circle = 2r = 2 × OQ = 2 × 30 = 60

∴ Diameter = 60

Given that AP and AQ are tangents to the circle.

A line l touches the circle at R and intersects AP and AQ in B and C respectively. And AB = c, BC = a, CA = b.

We have to prove that

(1) AP + AQ = a + b + c

(2) AB + BR = AC + CR = AP = AQ =

Proof:

By theorem,

AP = AQ, BP = BR and CQ = CR … (1)

(1) AP + AQ = (AB + BP) + (AC + CQ)

= (AB + BR) + (AC + CR) [From (1)]

= AB + AC + (BR + CR)

Since B – R – C,

⇒ AP + AQ = AB + AC + BC

= c + b + a

∴ AP + AQ = a + b + c … (2)

(2) AB + BR = AB + BP [From (1)]

= AP [∵ A – B – P]

= AQ [From (1)]

= AC + CQ [∵ A – C – Q]

= AC + CR [From (1)]

∴ AB + BR = AC + CR = AP = AQ … (3)

From (2),

⇒ AP + AQ = a + b + c

From (1),

⇒ AQ + AQ = a + b + c

⇒ 2AQ = a + b + c

⇒ AQ =

From (3),

∴ AB + BR = AC + CR = AP = AQ =

Hence proved.

Let l be a tangent to the circle having centre O. l touches the circle at P. Let m be the perpendicular line to l from P.

We have to prove that m passes through O i.e. O ∈ m.

Proof:

If O m, then we can find such M m that O and M are in the same half plane of l.

T ∈ l is a distinct point from P.

∴ ∠MPT = 90° and ∠OPT = 90°

M and O are points of the same half plane so this is impossible.

Thus, our assumption is wrong.

∴ O ∈ m

∴The perpendicular drawn to a tangent to the circle at the point of contact of the tangent passes through the centre of the circle.

Hence proved.

Let PA and PB be tangents to the circle drawn from point P which is in the exterior of circle (O, r). Given A and B are on the circle.

We have to prove that OP ⊥ AB and OP bisects AB.

Proof:

Here PA and PB are tangents drawn to the circle from an exterior point P.

∴ OP intersects AB at C.

Also given that A and B are on the circle.

We know that the tangents drawn to a circle from a point in the exterior of the circle are congruent.

∴ PA = PB

⇒ OP = OP [common]

⇒ OA = OB [radii of circle]

∴ By SSS theorem,

ΔOAP ≅ ΔOBP

Then, ∠AOP = ∠BOP

⇒ ∠AOC = ∠BOC [C ∈ OP]

⇒ OA = OB [radii of circle]

⇒ OC = OC [common]

∴ By SAS theorem,

ΔAOC ≅ ΔBOC

∴ AC = BC and ∠ACO = ∠BCO = 90°

Now, as C ∈ OP,

⇒ OP bisects AB.

Also AC ⊥ OC and BC ⊥ OC

⇒ OP ⊥ AB [∵ A – C – B]

∴ OP ⊥ AB and OP bisects AB.

Hence proved.

Given that PT and PR are tangents drawn to circle (O, r) from point P lying in the exterior of the circle and T and R are their points of contact respectively.

We have to prove that m∠TPR = 2m∠OTR

Proof:

By theorem,

PT ≅ PR

We know that angles opposite to congruent sides are equal.

∴ ∠PTR = ∠PRT

We know that sum of all angles in a triangle is 180°.

In ΔPTR,

⇒ ∠PTR + ∠PRT + ∠TPR = 180°

⇒ ∠PTR + ∠PTR + ∠TPR = 180°

⇒ 2∠PTR + ∠TPR = 180°

⇒ 2∠PTR = 180° – ∠TPR

⇒ ∠PTR = 90° – 1/2 ∠TPR

∴ 1/2 ∠TPR = 90° – ∠PTR … (1)

Then, OT ⊥ PT,

⇒ ∠OTP = 90°

⇒ ∠OTR + ∠PTR = 90°

⇒ ∠OTR = 90° – ∠PTR … (2)

From (1) and (2),

⇒ 1/2 ∠TPR = ∠OTR

⇒ ∠TPR = 2∠OTR

∴ m∠TPR = 2m∠OTR

Hence proved.

Given AB is a chord of circle (O, 5) such that AB = 8.

Let PR = x.

Since OP is a perpendicular bisector of AB,

AR = BR = 4

Consider ΔORA where ∠R = 90°,

By Pythagoras Theorem,

⇒ OA² = OR² + AR²

⇒ 5² = OR² + 4²

⇒ OR² = 25 – 16 = 9

∴ OR = 3

⇒ OP = PR + RO = x + 3

Consider ΔARP where ∠R = 90°,

⇒ PA² = AR² + PR²

⇒ PA² = 4² + x² = 16 + x² … (1)

Consider ΔOAP where ∠A = 90°,

⇒ PA² = OP² – OA²

= (x + 3)² – (5)²

= x² + 9 + 6x – 25

= x² + 6x – 16 … (2)

From (1) and (2),

⇒ 16 + x² = x² + 6x – 16

⇒ 6x = 32

⇒ x =

From (1),

⇒ PA² = 16 + x²

= 16 + () ²

= 16 +

=

=

∴ PA =

Given P lies in the exterior of circle (O, 5) such that OP = 13.

Let OR = x.

Consider ΔOAP where ∠A = 90°,

By Pythagoras Theorem,

⇒ OP² = OA² + AP²

⇒ 13² = 5² + AP²

⇒ AP² = 169 – 25 = 144

∴ AP = 12

Consider ΔORA where ∠R = 90°,

⇒ AR² = OA² – OR²

⇒ AR² = 5² – x² = 25 – x² … (1)

Consider ΔARP where ∠R = 90°,

⇒ AR² = AP² – PR²

= (12)² – (13 – x)²

= 144 – (13 + x² – 26x)

= – x² + 26x + 131 … (2)

From (1) and (2),

⇒ 25 – x² = – x² + 26x + 131

⇒ 26x = 50

⇒ x =

From (1),

⇒ AR² = 25 – x²

= 25 – () ²

= 25 –

=

=

∴ AR =

But AB = 2AR,

⇒ AB = 2 ()

∴ AB = = 9.23

Given that a circle touches the sides BC, CA and AB of ΔABC at points D, E and F respectively.

Let BD = x, CE = y and AF = z.

We have to prove that area of ΔABC =

Proof:

BD and BF are tangents drawn from B. And D and F are points of contact.

∴ BD = BF = x

Similarly for tangents CE and CD drawn by C and tangents AE and AF drawn from A,

CE = CD = y and AF = AE = z

Sides of ΔABC,

⇒ AB = c = AF + BF = z + x … (1)

⇒ BC = a = BD + DC = x + y … (2)

⇒ CA = b = CE + AE = y + z … (3)

In ΔABC, 2s = AB + BC + AC

= (z + x) + (x + y) + (y + z)

= 2 (x + y + z)

∴ s = x + y + z … (4)

We know that area of ΔABC =

Area =

=

=

Hence proved.

Given that ΔABC is an isosceles triangle in which AB ≅ AC and a circle touching all the three sides of ΔABC touches BC at D.

We have to prove that D is the midpoint of BC.

Proof:

We know that if a circle touches all sides of a triangle, the lengths of tangents drawn from each vertex are equal.

∴ AE = AF, BD = BF and CD = CE … (1)

Consider AB = AC,

Subtracting AF from both sides,

⇒ AB – AF = AC – AF

From (1),

⇒ AB – AF = AC – AE

Since A – F – B and A – E – C,

⇒ BF = CE

From (1),

⇒ BD = CD

Since B – D – C and BD = CD,

D is the midpoint of BC.

Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.

In the figure,

ID = IE = IF = r

Since ΔABC is a right angled triangle, ∠B = 90°.

Also ID ⊥ BC and AB ⊥ BC.

∴ ID || AB and ID || FB

Similarly, IF || BD

∴ IFBD is a parallelogram.

∴ FB = ID = r and BD = IF = r … (1)

∴ Parallelogram IFBD is a rhombus.

Since ∠B = 90°, parallelogram IFBD is a square.

By Pythagoras Theorem,

⇒ AC² = AB² + BC²

= 24² + 7²

= 576 + 49

= 625

∴ AC = 25

⇒ AB + BC + AC = 24 + 7 + 25

⇒ AF + FB + BD + DC + AC = 56

⇒ AE + r + r + CE + AC = 56

⇒ 2r + (AE + CE) + AC = 56

⇒ 2r + 2AC = 56

⇒ 2r + 2(25) = 56

⇒ r + 25 = 28

⇒ r = 3

∴ The radius of circle is 3.

Given that a circle touches all the three sides of a right angled ΔABC.

We have to prove that radius of a circle =

Proof:

Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.

In the figure,

ID = IE = IF = r

Since ΔABC is a right angled triangle, ∠B = 90°.

Also ID ⊥ BC and AB ⊥ BC.

∴ ID || AB and ID || FB

Similarly, IF || BD

∴ IFBD is a parallelogram.

∴ FB = ID = r and BD = IF = r … (1)

∴ Parallelogram IFBD is a rhombus.

Since ∠B = 90°, parallelogram IFBD is a square.

Now AE = AF

⇒ AE = AB – FB

= AB – r [From (1)] … (2)

And CE = CD

⇒ CE = BC – BD

= BC – r [From (1)] … (3)

Now, AC = AE + CE,

⇒ AC = AB – r + BC – r

⇒ AC = AB + BC – 2r

⇒ 2r = AB + BC – AC

⇒ r =

∴ The radius of a circle is .

Given that in â ABCD, ∠D = 90°. BC = 40, CD = 30 and BP = 25

We know that tangents drawn to a circle are perpendicular to the radius of the circle.

⇒ ∠ORD = ∠OSD = 90°

Given ∠D = 90° and OR = OS = radius.

∴ ORDS is a square.

We know that the tangents drawn to a circle from a point in the exterior of the circle are congruent.

∴ BP = BQ, CQ = CR and DR = DS.

Consider BP = BQ,

⇒ BQ = 25 [BP = 25]

⇒ BC – CQ = 40

⇒ CQ = 40 – 25 = 15 [BC = 40]

Consider CQ = CR,

⇒ CR = 15

⇒ CD – DR = 15

⇒ DR = 30 – 15 = 15 [CD = 30]

But ORDS is a square.

∴ OR = DR = 15

∴ Radius of circle OR is 15.

Given are two concentric circles.

Let two chords PQ and RS of the circle with larger radius touch the circle with smaller radius.

We have to prove that PQ ≅ RS.

Proof:

Let two chords PQ and RS of the circle with larger radius touch the circle with smaller radius at points M and N respectively.

PQ and RS are tangents to the circle with smaller radius,

∴ OM = ON = radius of smaller circle.

Thus, chords AB and CD are equidistant from the centre of the circle with larger radius.

∴ PQ = RS

∴ PQ ≅ RS

∴All chords of the circle with larger radius which touch the circle with smaller radius are congruent.

Hence proved.

We know that if a circle touches all the sides of a quadrilateral, then AB + CD = BC + DA

Given AB = 5, BC = 8, CD = 6

⇒ 5 + 6 = 8 + DA

⇒ 11 = 8 + DA

⇒ DA = 3

∴ AD = 3

Given that a circle touches all the sides of ABCD and AB is the largest side.

We have to prove that CD is the smallest side.

Proof:

The circle touches all sides of ABCD.

∴ AB + CD = BC + DA … (1)

Given AB is the largest side.

⇒ AB > BC

∴ AB = BC + m

From (1),

⇒ BC + m + CD = BC + DA

⇒ CD + m = DA

∴ CD < DA

Hence CD is smaller than DA. … (2)

But AB is the largest side.

⇒ AB > DA

∴ AB = DA + n

From (1),

⇒ DA + n + CD = BC + DA

⇒ CD + n = BC

∴ CD < BC

Hence CD is smaller than BC. … (3)

AB is largest side, so CD is smaller than AB. … (4)

From (2), (3) and (4), CD is the smallest side of ABCD.

Given P lies in the exterior of a circle having centre O and PQ is a tangent.

∴ OQ ⊥ PQ

Also given OP = 25 and OQ = 24.

Consider ΔOQP,

∠OQP = 90°

By Pythagoras Theorem,

⇒ OP² = OQ² + PQ²

⇒ 25² = 24² + PQ²

⇒ PQ² = 625 – 576

= 49

∴ PQ = 7

Given P is in exterior of circle (O, 15) and a tangent from P touches the circle at T.

Thus, ∠OTP = 90°

By Pythagoras Theorem,

⇒ OP² = OT² + TP²

⇒ OP² = 15² + 8²

= 225 + 64

= 289

∴ OP = 17

Given PA and PB touch circle (O, 15) at A and B and m ∠AOB = 80°.

Here, ΔPOA and ΔPOB are congruent right angled triangle.

⇒ ∠BOP = 1/2 ∠AOB

= 1/2 × 80°

= 40°

In right angled ΔOBP,

We know that sum of angles in a triangle is 180°.

⇒ ∠BOP + ∠B + ∠OPB = 180°

⇒ 40 + 90 + ∠OPB = 180°

⇒ 130 + ∠OPB = 180°

∴ ∠OPB = 50°

Given OP = 13 and PQ = 5

In right angled ΔOQP,

By Pythagoras Theorem,

⇒ OP² = OQ² + PQ²

⇒ 13² = r² + 5²

⇒ r² = 169 – 25 = 144

∴ r = 12

Diameter of circle = 2r

= 2 (12)

= 24

∴ Diameter = 24

Given in ΔABC, AB = 3, BC = 4, AC = 5.

By Pythagoras Theorem,

⇒ AC² = AB²+ BC²

⇒ 5² = 3² + 4²

⇒ 5² = 9 + 16

⇒ 5² = 5²

∴ ΔABC is a right angled triangle and ∠B is a right angle.

We know that the radius of the circle touching all the sides is .

⇒ The required radius of circle =

=

= 1

Given ∠OPB = 30° and OP = 10

In right angled ΔOBP,

Consider sin30° =

⇒ 1/2 =

⇒ OB = 5

∴ Radius of circle = OB = 5

In right angled ΔOBP,

Given ∠OBP = 30°

⇒ ∠BOP + ∠OPB + ∠B = 180°

⇒ ∠BOP + 30° + 90° = 180°

⇒ ∠BOP = 180° – 120° = 60°

∴ ∠BOP = 60°

Now, ∠AOB = 2 ∠BOP

= 2 (60°)

= 120°

∴ ∠AOB = 120°

Given chord of circle (O, 5) touches circle (O, 3).

Radius of smaller circle OM = 3 and radius of bigger circle OB = 5

In right angled ΔOMB,

⇒ OB² = OM² + MB²

⇒ 5² = 3² + MB²

⇒ MB² = 25 – 9 = 16

⇒ MB = 4

Length of chord AB = 2 (MB) = 2 (4) = 8