Circles

Given: Diameter of the circle = d = AD = 34 cm

∴ Radius of the circle = r = d/2 = AO = 17 cm

Length of chord AB = 30 cm

Since the line drawn through the center of a circle to bisect a chord is perpendicular to the chord, therefore AOL is a right angled triangle with L as the bisector of AB.

∴ AL = 1/2(AB) = 15 cm

In right angled triangle AOB, by Pythagoras theorem, we have:

(AO)² = (OL)² + (AL)²

⇒ (17)² = (OL)² + (15)²

⇒ (OL)² = (17)² - (15)²

⇒ (OL)² = 289 – 225

⇒ (OL)² = 64

Take square root on both sides:

⇒ (OL) = 8

∴ The distance of AB from the center of the circle is 8 cm.

Given:

Radius of the circle = r = AO = 5 cm

Length of chord AB = 8 cm

Since the line drawn through the center of a circle to bisect a chord is perpendicular to the chord, therefore AOC is a right angled triangle with C as the bisector of AB.

∴ AC = 1/2(AB) = 8/2 = 4 cm

In right angled triangle AOC, by Pythagoras theorem, we have:

(AO)² = (OC)² + (AC)²

⇒ (5)² = (OC)² + (4)²

⇒ (OC)² = (5)² - (4)²

⇒ (OC)² = 25 – 16

⇒ (OC)² = 9

Take square root on both sides:

⇒ (OC) = 3

∴ The distance of AC from the center of the circle is 3 cm.

Now, OD is the radius of the circle, ∴ OD = 5 cm

CD = OD – OC

CD = 5 – 3

CD = 2

∴ CD = 2 cm

Given: AB = 12 cm, BC = 16 cm, AB ⊥ BC.

Therefore, the circle passing through the points A, B and C has AC as its diameter.

So, from the figure, ABC is a right angled trianlge.

By Pythagoras theorem:

(AC)² = (CB)² + (AB)²

⇒ (AC)² = (16)² + (12)²

⇒ (AC)² = 256 + 144

⇒ (AC)² = 400

Take square root on both sides:

⇒ (AC) = 20

∴ Diameter of the circle = 20 cm

Thus, radius of the circle = Diameter/2 = 20/2 = 10 cm

∴ Radius of the circle = 10 cm

Given: ∠ABC = 20°

By theorem “The angle subtended by an arc at the center of a circle is twice the angle subtended by it at remaining part of the circle”, we have:

∠AOC = 2 × ∠ABC

= 2 × 20°

= 40°

∴ ∠AOC = 40°

Given: AOB is the diameter of the circle.

AC = BC

⇒ ∠ABC = ∠BAC = x (say) (∵ angles opposite to equal sides are equal)

Also, diameter subtends a right angle to the circle,

∴ ∠ACB = 90°

Now, by angle sum property of a triangle, sum of all angles of a triangle is 180°.

∴ ∠CAB + ∠ABC + ∠ACB = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 90°

⇒ x = 45°

∴ ∠CAB = ∠ ABC = 45°

In triangle AOB,

AO = OB = Radius

∴ ∠OAB = ∠OBA = 40° (∵ angles opposite to equal sides are equal)

Using the angle sum property of triangle, sum of all angles of a triangle is 180°,

∴ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ 40° + 40° + ∠AOB = 180°

⇒ ∠AOB = 180° - 40° - 40°

⇒ ∠AOB = 100°

By theorem “The angle subtended by an arc at the center of a circle is twice the angle subtended by it at remaining part of the circle”, we have:

∠AOB = 2 × ∠ACB

∠ACB = ∠AOB/2

= 100°/2

∴ ∠ACB = 50°

Given: ∠DAB = 60°, ∠ABD = 50°

Using the theorem “angles in the same segment of the circle are equal”:

∴ ∠ADB = ∠ACB ……………… (1)

In triangle ABD, using the angle sum property of triangle, we have:

∠ABD + ∠ADB +∠DAB = 180°

⇒ 50° + ∠ADB + 60° = 180°

⇒ ∠ADB = 180° - 50° - 60°

⇒ ∠ADB = 70°

∴ From equation (1), we have:

∠ACB = 70°

Given: ABCD is a cyclic quadrilateral.

∠ADC = 140°

Since sum of opposite angles of a cyclic quadrilateral is 180°,

∴ ∠ADC + ∠ABC = 180°

⇒ 140° + ∠ABC = 180°

⇒ ∠ABC = 180° - 140°

⇒ ∠ABC = 40°

Since, diameter subtends a right angle to the circle,

∴ ∠ACB = 90°

Now, in triangle ACB; by angle sum property of a triangle, sum of all angles of a triangle is 180°.

∴ ∠CAB + ∠ABC + ∠ACB = 180°

⇒ ∠CAB + 40° + 90° = 180°

⇒ ∠CAB = 180° - 90° - 40°

⇒ ∠CAB = 50°

∴ ∠CAB = ∠50°

Given: ∠BAO = 60°

Since, OA and OB are radius of the circle, therefore OA = OB

⇒ ∠ ABO = ∠BAO = 60° (∵ given that ∠BAO = 60°)

Since angle in the same segment are equal, therefore

∠ABC = ∠ADC (angles in the same segment AC are equal)

∠ABC = ∠ABO = 60°

∴ ∠ADC = 60°

Given: ∠AOB = 90°, ∠ABC = 30°

OA and AB are radius and are equal,

∴ ∠OAB = ∠OBA = x (say) (∵ angles opposite to equal sides are equal)

In triangle OAB, using the angle sum property, sum of all angles of triangle is 180°.

∴ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ 90° + x + x = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = 45°

∴ ∠OAB = ∠OBA = 45°

Since angles in the same segment are equal, therefore

∠ACB = ∠OAB = 45° (∵ ∠OAB = 45° and these angles lie in the same segment CB)

Now, in triangle CAB, using the angle sum property of triangle, sum of all angles is 180°.

∴ ∠CAB + ∠ABC + ∠BCA = 180°

⇒ ∠CAB + 30° + 45° = 180°

⇒ ∠CAB = 180° - 30° - 45°

⇒ ∠CAB = 105°

⇒ ∠CAO + ∠OAB = 105° (∵ ∠CAB = ∠CAO + ∠OAB)

⇒ ∠CAO + 45° = 105°

⇒ ∠CAO = 105° - 45°

⇒ ∠CAO = 60°

∴ ∠CAO = 60°

TRUE

Given that AB and AC are chords that are at a distance of 4 cm from center of a circle.

Since, chords that are equidistant from the center of a circle are equal in length, therefore AB = CD.

FALSE

Let AB and AC be the chord of the circle with center O on the opposite side of OA.

Consider the triangles AOC and AOB:

AO = AO (Common side in both triangles)

OB = OC (Both OB and OC are radius of circle)

But we can’t show that either the third side of both triangles are equal or any angle is equal. Therefore ΔAOB is not congruent to ΔAOC.

∴ ∠OAB ≠ ∠OAC.

TRUE

Let the congruent circles with centers O and O’ intersect at A and B. Join AB, O’A, O’B, OA and OB.

By joining the points, we obtain two triangles, namely OAB and O’AB.

Since both the circles are congruent, therefore in ΔOAB and ΔO’AB, we have:

OA = O’A (Both circles have same radius as the circles are congruent.)

OB = O’B (Both circles have same radius as the circles are congruent.)

AB = AB (Common)

∴ By SSS congruence rule, ΔOAB = ΔO’AB

∴ By CPCT, ∠AOB = ∠AO’B

FALSE

A circle can pass through only two collinear points, not through three collinear points.

TRUE

Let AB be of length 6 cm.

Then take the midpoint of AB which will be at 3 cm.

Let the point be C.

Consider C as the center of circle and take the radius as 3 cm and draw a circle. That circle will pass through A and B both as the radius is 3 cm and AC = CB = 3cm = Radius of the circle and AB will serve as the diameter of the circle.

TRUE

Let AB be the diameter of the circle with center O and C be any point on circle.

Since, diameter subtends a right angle to the circle,

∴ ∠ACB = 90°

Now, in right angled triangle ACB, by Pythagoras theorem, we have:

(AB)² = (AC)² + (AB)²

Thus, the statement is true.

FALSE

Given: ∠A = 90°, ∠B = 70°, ∠C = 95°, ∠D = 105°

Since sum of opposite angles of a cyclic quadrilateral is 180°;

∴ ∠A + ∠C = 90° + 95° = 185°, which can’t be true.

Also, ∠B + ∠D = 70° + 105° = 175°, which can’t be true.

Thus, there can’t exist such a cyclic quadrilateral.

FALSE

Given: There can be many points D such that ∠BDC = 60° but each such point cannot be the center of the circle that passes through A, B and C.

TRUE

In geometry, a set of points are said to be concyclic if they lie on a common circle.

Given: ∠BAC = ∠BDC = 45°

Since, angles in the same segment are equal, therefore A, B, C and D are concyclic.

Let AOB be the diameter of the circle.

Given: ∠ADC = 120°

Firstly, join CB.

Then, we have a cyclic quadrilateral ABCD.

Since sum of opposite angles of cyclic quadrilateral is 180°, therefore

∠ADC + ∠ABC = 180°

⇒ 120° + ∠ABC = 180°

⇒ ∠ABC = 180° - 120°

⇒ ∠ABC = 60°

Now join AC.

Also, diameter subtends a right angle to the circle,

∴ In ΔABC, ∠ACB = 90°

Now, by angle sum property of a triangle, sum of all angles of a triangle is 180°.

∴ ∠CAB + ∠ABC + ∠ACB = 180°

⇒ ∠CAB + 60° + 90° = 180°

⇒ ∠CAB = 180° - 90° - 60°

⇒ ∠CAB = 30°

∴ The given statement is true.

Given: AXB ≅ CYD.

Since the two arcs of a circle are congruent, then their corresponding arcs are also equal.

So, we have chord AB = chord CD.
Hence. AB : CD = 1:1

Let O be the center of the circle, AB be the chord and PQ be the perpendicular bisector of AB which intersects AB at point M.

The perpendicular bisector of the chord AB always passes through the center O.

Join PA and PB so that we have two triangles PAM and PBM.

In ΔPAM and ΔPBM:

AM = BM (M is the midpoint of AB ∵ perpendicular

bisector of AB intersects at M)

∠PMA = ∠PMB = 90°

PM = PM (common)

∴ ΔPAM ≅ ΔPBM (SAS congruence rule)

∴ AP = BP

⇒ arc PXA = arc PYB (by CPCT)

A set of lines or curves are said to be concurrent if they all intersect at the same point.

Let A, B, C be points on a circle. Draw perpendicular bisector of AB and AC which meet at point O.

Join OA, OB and OC.

We need to prove that the perpendicular bisector of BC also passes through O (if so, then perpendicular bisectors of AB, BA and CA are concurrent as they all will intersect at the same point O).

So, in ΔOEB and ΔOEA:

AE = BE (∵ E is the perpendicular bisector of AB)

∠AEO = ∠BEO = 90°

OE = OE (common)

∴ ΔOEB ≅ ΔOEA (by SAS congruence rule)

∴ OA = OB (By CPCT)

Similarly, ΔOFA ≅ ΔOFB (by SAS congruence rule)

∴ OA = OC (By CPCT)

So, OA = OB = OC = x (say)

Construct a perpendicular line from O to the line BC which intersect line BC at M and join them.

So, in ΔOMB and ΔOMC:

OB = OC (proved above)

OM = OM (common)

∠OMB = ∠OMC = 90°

∴ ΔOEB ≅ ΔOEA (by RHS congruence rule)

⇒ BM = MC (by CPCT)

∴ M is the perpendicular bisector of BC and hence OL, ON and OM are concurrent.

Let AB and BC be equal chords of circle with center O.

Draw the angle bisector AD of ∠BAC.

Join BC, meeting AD at M.

In triangle DAM and CAM:

AB = BC (given that the chords are equal)

∠BAM = ∠CAM (∵ AD is angle bisector of A)

AM = AM (common side)

∴ ΔABD = ΔACD (by SAS congruence rule)

∴ BM = CM and ∠AMB = ∠AMC = x (say) (by CPCT) ………(1)

But ∠AMB + ∠AMC = 180° ………(2)

From equation (1) and (2), we have:

x + x = 180°

⇒ 2x = 1280°

⇒ x = 90°

⇒ ∠AMB = ∠AMC = 90°

Therefore, AD is the perpendicular bisector of chord BC, but the perpendicular bisector of any chord always passes from the center of the circle.

∴ AD passes through the center O of the circle.

Thus, O lies on the angle bisector of the angle BAC.

Hence, proved.

Let AB and CD be the chords of the circle with center O.

Let L and M be the mid points of AB and CD respectively and PQ is the line passing through these midpoints and the center of the circle. Therefore PQ is the diameter of the circle.

Since M is the midpoint of CD,

∴ OM ⊥ CD (because line joining center to the midpoint of a chord is always perpendicular to the chord.)

⇒ OMD = 90°

Similarly, L is the midpoint of AB,

∴ OL ⊥ AB (because line joining center to the midpoint of a chord is always perpendicular to the chord.)

⇒ OLA = 90°

But ∠OLA and ∠OMD are alternate angles, so AB ∥ CD.

Hence, proved.

Join CA and BD.

Since arc DC subtends ∠DAC at the center and ∠CAB at point B in the remaining part of the circle, we have:

∠DAC = 2∠CBD ……………(1)

(Reason: In a circle, angle subtended by an arc at the center is twice the angle subtended by it at any other point in the remaining part of the circle.)

Similarly, arc BC subtends ∠CAB at the center and ∠CDB at point D in the remaining part of the circle, we have:

∠CAB = 2∠CDB ……………(2)

(Reason: In a circle, angle subtended by an arc at the center is twice the angle subtended by it at any other point in the remaining part of the circle.)

From equation (1) and (2), we have:

∠DAC + ∠CAB = 2∠CDB + 2∠CBD

⇒ ∠BAD = 2(∠CDB + ∠CBD)

⇒ 2(∠CDB + ∠CBD) = 1/2 (∠BAD)

Given: O is the circumcenter of the triangle ABC and D is the midpoint of BC.

To prove: ∠BOD = ∠A

Join OB and OC.

In ΔOBD and ΔCD:

OD = OD (common side)

DB = Dc (D is the midpoint of BC)

OB = OC (Both are radius of the circle)

By SSS congruence rule, ΔOBD ≅ ΔOCD.

∴ ∠BOD = ∠COD = x (say) (By CPCT)

Since, angle subtended by an arc at the center of the circle is twice the angle subtended by it at any other point in the remaining part of the circle, we have:

2∠BAC = ∠BOC

⇒ 2∠BAC = ∠BOD + ∠DOC

⇒ 2∠BAC = x + x

⇒ 2∠BAC = 2x

⇒ ∠BAC = x

⇒ ∠BAC = ∠BOD

Hence, proved.

Given: ACB and ADB are two right triangles.

To Prove: ∠BAC = ∠BDC

Since ACB and ADB are right angled triangles, therefore

∠C + ∠D = 90° + 90°

= 180°

Therefore ADBC is a cyclic quadrilateral. (∵ Sum of opposite angles of a cyclic quadrilateral is 180°.)

Also, ∠BAC and ∠BDC lie in the same segment BC and angles in the same segment of a circle are equal.

∴ ∠BAC = ∠BDC.

Hence Proved.

In ΔAOB, OA = OB (both are radius of the circle)

∴ ∠OBA = ∠OAB (angle opposite to equal sides are equal)

Using the angle sum property in ΔAOB, sum of all angles of the triangle is 180°, we have:

∠OAB + ∠AOB +∠OBA = 180°

⇒ ∠OAB +90° + ∠OAB = 180°

⇒ 2∠OAB = 180° - 90°

⇒ 2∠OAB = 90°

⇒ ∠OAB = 45°

Now, in ΔAOC,

OA = OC (both are radius of the circle)

∴ ∠OCA = ∠OAC (angle opposite to equal sides are equal)

Using the angle sum property in ΔAOB, sum of all angles of the triangle is 180°, we have:

∠OAC + ∠AOC +∠OCA = 180°

⇒ ∠OAC +150° + ∠OAC = 180°

⇒ 2∠OAC = 180° - 150°

⇒ 2∠OAC = 30°

⇒ ∠OAC = 15°

Now, ∠BAC = ∠OAB + ∠OAC

= 45° + 15°

= 60°

∴ ∠BAC = 60°

Let O be the center of BC.

Since CN is perpendicular on AB, therefore ΔBNC is right angled triangle.

Therefore the circle passing through B, N and C will have midpoint of BC as center and OB be the radius. …………… (1)

Similarly, BM is perpendicular on AC, therefore ΔBMC is right angled triangle.

Therefore the circle passing through B, M and C will have midpoint of BC as center and OB be the radius. …………… (2)

From (1), we get a circle passing through B, N and C which is centered at O and has radius OB.

From (2), we get a circle passing through B, M and C which is centered at O and has radius OB.

Since from a fixed point and fixed radius, only one circle can be drawn. Therefore, same circle will pass through the four points B, M, N and C.

Therefore; B, N, M and C are concyclic.

Given: ABC is an isosceles triangle such that AB = AC and ED ∥ BC.

To prove: Quadrilateral BCDE is cyclic, ie. sum of opposite angles is 180°.

In ΔABC,

AB = AC (Equal sides of the isosceles triangle)

⇒ ∠ABC = ∠ACB (angles opposite to equal sides are equal)

As ED ∥ BC, therefore,

∠ADE = ∠ACB (corresponding angles)

Adding ∠EDC on both sides, we get:

∠ADE + ∠EDC = ∠ACB + ∠EDC

⇒ 180° = ∠ACB + ∠EDC

⇒ 180° = ∠ABC + ∠EDC (∵ ∠ACB = ∠ABC)

Sum of opposite angles of quadrilateral is 180°.

⇒ Quadrilateral BCDE is cyclic.

Hence, proved.

Let ABCD be a cyclic quadrilateral with AD = BC.

We need to prove that AC = DB.

In triangles AOD and BOC:

AD = DB (given)

∠OAD = ∠OBC and ∠ODA = ∠OCB (angle subtended by the same segment are equal)

∴ ΔAOD ≅ ΔBOC (By ASA congruence rule)

⇒ ΔAOD + ΔDOC ≅ ΔBOC + ΔDOC (adding a similar quantity on both sides)

⇒ ΔADC ≅ ΔBCD

∴ AC = BD (by CPCT).

Hence, proved.

Let ABC be the triangle whose circumcenter is O.

∠OBC = ∠OCB = θ (opposite angles of equal sides

InΔBOC, using the angle sum property of tringle, sum of all angles is 180°, we have:
∠BOC +∠OBC +∠OCB = 180°
⇒∠BOC +θ + θ= 180°

⇒∠BOC = 180° -2θ

Also, in a circle, angle subtended by an arc at the center is twice the angle subtended by it at any other point in the remaining part of the circle.
∠BOC = 2∠BAC

⇒∠BAC = 1/2(∠BOC)

⇒∠BAC = 1/2(180° - 2θ)

⇒∠BAC = (90° - θ)

⇒∠BAC + θ = 90°

⇒∠BAC +∠OBC = 90°

Hence, proved.

Let AB be the chord of the circle with center O.

Given that AB = Radius of the circle.

Also, AO = BO = Radius

∴ ΔOAB is an equilateral triangle.

Thus, ∠AOB = ∠OBA = ∠OAB = 60°

Also, angle subtended by an arc at the center of the circle is twice the angle subtended by it at any other point in the remaining part of the circle.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = 1/2 (∠AOB)

⇒ ∠ACB = 1/2 (60°) = 30°

Given: ∠ADC = 130°

chord BC = chord BE

Since, ABCD is a cyclic quadrilateral, therefore sum of its opposite angles is 180°.

⇒ ∠ADC + ∠OBC = 180°

⇒ 130° + ∠OBC = 180°

⇒ ∠OBC = 50°

Now, in triangles BOC and BOE:

OB = OB (common)

BC = BE (given)

OC = OE (radius of the circle)

By SSS congruence rule, ΔBOC ≅ ΔBOE.

∴ ∠OBC = ∠OBE = 50° (By CPCT)

Now, ∠CBE = ∠CBO + ∠OBE

∠CBE = 50° + 50°

∠CBE = 100°

Given: ∠ACB = 40°

Since, the angle subtended by an arc at the center of a circle is twice the angle subtended by it at remaining part of the circle, therefore we have:

∠AOB = 2 × ∠ACB

= 2 × 40°

= 80°

Now, in triangle AOB, AO and BO are both radius of the circle.

Therefore, ∠OAB = ∠OBA = x (say) (angles opposite to equal sides are equal)

Using the angle sum property of triangle, sum of all angles of a triangle is 180°, we have:

∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 80° = 180°

⇒ 2x = 100

⇒ x = 50°

⇒ ∠OAB = 50°

Given: AB is the diameter.

ABCD is the cyclic quadrilateral.

∠ADC = 130°

Since ABCD is the cyclic quadrilateral, therefore sum of opposite angles is 180°.

⇒ ∠ADC + ∠ABC = 180°

⇒ 130° + ∠ABC = 180°

⇒ ∠ABC = 180° - 130°

⇒ ∠ABC = 50°

Now, AB is the diameter and the angle subtended to the circle by the diameter is a right angle.

∴ ∠ACB = 90°

In triangle ACB, sum of all angles of triangle is 180°.

∴ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 90° + 50° + ∠BAC = 180°

⇒ ∠BAC = 180° - 90° - 50°

⇒ ∠BAC = 40°

Firstly draw two circles with center O and O’ such that they intersect at A and B.

Draw a line PQ parallel to OO’.

In the circle with center O, we have:

OP and OB are the radii of the circle. PB is the chord with OM as its perpendicular bisector.

i.e. BM = MP ………………………(1)

In the circle with center O’, we have:

O’B and O’Q are the radii of the circle. BQ is the chord with O’N as its perpendicular bisector.

i.e. BN = NQ ………………………(2)

From (1) and (2), we have:

BM + BN = MP + NQ

⇒ (BM + BN) + (BM + BN) = (BM + BN) + (MP + NQ)

⇒ 2(BM + BN) = (BM + BN) + (MP + NQ)

⇒ 2(OO’) = (BM + MP) + (BN + NQ)

⇒ 2(OO’) = BP + BQ

⇒ 2(OO’) = PQ

Hence, proved.

Given: AOB is the diameter.

Join AE and ACDE is a cyclic quadrilateral.

∴ ∠ACD + ∠AED = 180° (sum of opposite angles of a cyclic quadrilateral is 180°) ………………(1)

As diameter subtend a right angle to the circle, therefore,

∠AEB = 90° ………………(2)

Adding equation (1) and (2), we have:

∠ACD + ∠AED + ∠AEB = 180° + 90°

⇒ ∠ACD + ∠BED = 270°

Given: ∠OAB = 30°, ∠OCB = 57°

In ΔOAB, AO = BO (Both are radii of the circle).

Thus ∠OAB = ∠OBA = 30° (angles opposite to equal sides are equal)

In ΔAOB, sum of all angles of a triangle is 180°.

∴ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ 30° + 30° + ∠AOB = 180°

⇒ ∠AOB = 180° - 30° - 30°

⇒ ∠AOB = 120° …………………………… (1)

Now, in triangle OBC, OC and OB are radius of the circle and are thus equal.

∴ ∠OBC = ∠OCB = 57° (angles opposite to equal sides are equal)

In ΔAOB, sum of all angles of a triangle is 180°.

∴ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ 57° + 57° + ∠BOC = 180°

⇒ ∠BOC = 180° - 57° - 57°

⇒ ∠BOC = 66° …………………………… (2)

Now, from equation (1), we have:

∠AOB = 120°

⇒ ∠AOC + ∠COB = 120°

⇒ ∠AOC + 66° = 120°

⇒ ∠AOC = 120° - 66°

⇒ ∠AOC = 54°

Therefore, ∠AOC = 54° and ∠BOC = 66°.