Circle — Exercise 14.1

Calculate the length of a chord which is at a distance 12 cm from the centre of a circle of radius 13 cm.

Answer

From figure,

AB is the chord which is at a distance 12 cm so,

OC = 12 cm and OA = radius = 13 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 13² = 12² + AC²

⇒ AC² = 13² - 12²

⇒ AC² = 169 - 144 = 25

⇒ AC = $\sqrt{25}$ = 5 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = 5 cm.

AB = AC + CB = 5 + 5 = 10 cm.

Hence, length of chord = 10 cm.

A chord of length 48 cm is drawn in a circle of radius 25 cm. Calculate its distance from the centre of the circle.

Answer

From figure,

AB is the chord and radius = OA = 25 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = $\dfrac{48}{2}$ = 24 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 25² = OC² + 24²

⇒ OC² = 25² - 24²

⇒ OC² = 625 - 576 = 49

⇒ OC = $\sqrt{49}$ = 7 cm.

Hence, chord is at a distance of 7 cm from the center.

A chord of length 8 cm is at a distance 3 cm from the centre of the circle. Calculate the radius of the circle.

Answer

From figure,

AB is the chord which is at a distance 3 cm from the center so OC = 3 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = $\dfrac{8}{2}$ = 4 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ OA² = 3² + 4²

⇒ OA² = 9 + 16

⇒ OA² = 25

⇒ OA = $\sqrt{25}$ = 5 cm.

Hence, radius = 5 cm.

Calculate the length of a chord which is at a distance 6 cm from the center of a circle of diameter 20 cm.

Answer

Diameter = 20 cm,

∴ Radius = $\dfrac{\text{Diameter}}{2}$ = 10 cm.

From figure,

AB is the chord which is at a distance 6 cm so,

OC = 6 cm and OA = radius = 10 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 10² = 6² + AC²

⇒ AC² = 10² - 6²

⇒ AC² = 100 - 36 = 64

⇒ AC = $\sqrt{64}$ = 8 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = 8 cm.

AB = AC + CB = 8 + 8 = 16 cm.

Hence, length of chord = 16 cm.

A chord of length 16 cm is at a distance 6 cm from the center of the circle. Find the length of chord of the same circle which is at a distance of 8 cm from the center.

Answer

From figure,

AB is the chord which is at a distance 6 cm from the center so OC = 6 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CB = AC = $\dfrac{16}{2}$ = 8 cm.

In right angle triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ OA² = 6² + 8²

⇒ OA² = 36 + 64

⇒ OA² = 100

⇒ OA = $\sqrt{100}$ = 10 cm.

Radius = 10 cm,

∴ OD = 10 cm.

From figure,

In right angle triangle ODF,

⇒ OD² = OF² + DF² (By pythagoras theorem)

⇒ DF² = OD² - OF²

⇒ DF² = 10² - 8²

⇒ DF² = 100 - 64 = 36

⇒ DF = $\sqrt{36}$ = 6 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord

∴ DE = DF + FE = 6 cm + 6 cm = 12 cm.

Hence, the length of chord which is at a distance of 8 cm from the center of the circle = 12 cm.

In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords, if they are on

(i) the same side of the centre

(ii) the opposite sides of the centre.

Answer

(i) From figure,

AB and CD are chords of length 8 cm and 6 cm, respectively.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AE = BE = $\dfrac{8}{2}$ = 4 cm and,

CF = FD = $\dfrac{6}{2}$ = 3 cm.

Radius = OA = OC = 5 cm.

In right angle triangle OAE,

⇒ OA² = AE² + OE² (By pythagoras theorem)

⇒ OE² = OA² - AE²

⇒ OE² = 5² - 4²

⇒ OE² = 25 - 16 = 9

⇒ OE = $\sqrt{9}$ = 3 cm.

In right angle triangle OCF,

⇒ OC² = CF² + OF² (By pythagoras theorem)

⇒ OF² = OC² - CF²

⇒ OF² = 5² - 3²

⇒ OF² = 25 - 9 = 16

⇒ OF = $\sqrt{16}$ = 4 cm.

Distance between two chords = EF = OF - OE = 4 - 3 = 1 cm.

Hence, distance between two chords = 1 cm.

(ii) From figure,

AB and CD are chords of length 8 cm and 6 cm respectively.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AE = BE = $\dfrac{8}{2}$ = 4 cm and,

CF = FD = $\dfrac{6}{2}$ = 3 cm.

Radius = OA = OC = 5 cm.

In right angle triangle OAE,

⇒ OA² = AE² + OE² (By pythagoras theorem)

⇒ OE² = OA² - AE²

⇒ OE² = 5² - 4²

⇒ OE² = 25 - 16 = 9

⇒ OE = $\sqrt{9}$ = 3 cm.

In right angle triangle OCF,

⇒ OC² = CF² + OF² (By pythagoras theorem)

⇒ OF² = OC² - CF²

⇒ OF² = 5² - 3²

⇒ OF² = 25 - 9 = 16

⇒ OF = $\sqrt{16}$ = 4 cm.

Distance between two chords = EF = OF + OE = 4 + 3 = 7 cm.

Hence, distance between two chords = 7 cm.

In figure (i) given below, O is the center of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the :

(i) radius of the circle

(ii) length of chord CD.

Answer

(i) Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AM = BM = $\dfrac{24}{2}$ = 12 cm.

In right angle triangle OAM,

⇒ OA² = OM² + AM² (By pythagoras theorem)

⇒ OA² = 5² + 12²

⇒ OA² = 25 + 144

⇒ OA² = 169

⇒ OA = $\sqrt{169}$ = 13 cm.

Hence, radius = 13 cm.

(ii) From figure,

OC = radius = 13 cm.

In right angle triangle OCN,

⇒ OC² = ON² + CN² (By pythagoras theorem)

⇒ CN² = OC² - ON²

⇒ CN² = 13² - 12²

⇒ CN² = 169 - 144 = 25

⇒ CN = $\sqrt{25}$ = 5 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ ND = CN = 5 cm.

CD = CN + ND = 10 cm.

Hence, CD = 10 cm.

In the figure (ii) given below, CD is the diameter which meets the chord AB in E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.

Answer

Given,

AB = 8 cm, EC = 3 cm

Let radius OB = OC = r

OE = (r - 3) cm.

Since, chord AB is bisected by OE so OE ⊥ AB (As straight line drawn from the centre of circle to bisect a chord, is perpendicular to it.)

Now in right ∆OBE,

⇒ OB² = BE² + OE²

⇒ r² = 4² + (r – 3)²

⇒ r² = 16 + r² – 6r + 9

⇒ r² - r² + 6r = 16 + 9

⇒ 6r = 25

⇒ r = $\dfrac{25}{6} = 4\dfrac{1}{6}$ cm.

Hence, radius = $4\dfrac{1}{6}$ cm.

In the adjoining figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

Answer

In the figure, chords AB ∥ CD and O is the centre of the circle.

Radius of the circle = 15 cm

Length of AB = 24 cm and CD = 18 cm.

Join OA and OC.

AB = 24 cm and OM ⊥ AB.

∴ AM = MB = $\dfrac{24}{2}$ = 12 cm (As perpendicular to a chord from the center of the circle bisects it)

In right angle triangle OAM,

⇒ OA² = OM² + AM² (By pythagoras theorem)

⇒ OM² = OA² - AM²

⇒ OM² = 15² - 12²

⇒ OM² = 225 - 144

⇒ OM² = 81

⇒ OM = $\sqrt{81}$ = 9 cm.

Similarly ON ⊥ CD

CN = ND = $\dfrac{18}{2}$ = 9 cm

Similarly In right ∆CNO,

⇒ OC² = ON² + CN² (By pythagoras theorem)

⇒ ON² = OC² - CN²

⇒ ON² = 15² - 9²

⇒ ON² = 225 - 81

⇒ ON² = 144

⇒ ON = $\sqrt{144}$ = 12 cm.

MN = OM + ON = 9 + 12 = 21 cm.

Hence, MN = 21 cm.

AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm respectively. If the chords lie on the same side of the centre and the distance between them is 3 cm, find the diameter of the circle.

Answer

Let OE = x cm.

From figure,

In right angle triangle OCF,

⇒ OC² = OF² + CF² (By pythagoras theorem)

⇒ OC² = (x + 3)² + 2²

⇒ OC² = x² + 9 + 6x + 4

⇒ OC² = x² + 6x + 13

Since, radius = OA = OC.

∴ OA² = OC² = x² + 6x + 13.

In right angle triangle OAE,

⇒ OA² = OE² + AE²

⇒ x² + 6x + 13 = x² + 5²

⇒ x² - x² + 6x = 25 - 13

⇒ 6x = 12

⇒ x = $\dfrac{12}{6}$ = 2 cm.

⇒ OC² = x² + 6x + 13

⇒ OC² = 2² + 6(2) + 13

⇒ OC² = 4 + 12 + 13

⇒ OC² = 29

⇒ OC = $\sqrt{29}$ cm.

Diameter = 2 × radius = 2 × $\sqrt{29} = 2\sqrt{29}$ cm.

Hence, diameter = $2\sqrt{29}$ cm.

ABC is an isosceles triangle inscribed in a circle. If AB = AC = $12\sqrt{5}$ cm and BC = 24 cm, find the radius of the circle.

Answer

From figure,

OA = radius = r cm.

BD = DC = $\dfrac{24}{2}$ = 12 cm (As perpendicular to a chord from the center of the circle bisects it)

In right angle triangle ABD,

$\Rightarrow AB^2 = BD^2 + AD^2 \\[1em] \Rightarrow (12\sqrt{5})^2 = 12^2 + AD^2 \\[1em] \Rightarrow 720 = 144 + AD^2 \\[1em] \Rightarrow AD^2 = 720 - 144 \\[1em] \Rightarrow AD^2 = 576 \\[1em] \Rightarrow AD = \sqrt{576} = 24 \text{ cm}.$

OD = AD - OA = (24 - r) cm.

In right angle triangle OBD,

⇒ OB = radius = r cm.

⇒ OB² = OD² + BD²

⇒ r² = (24 - r)² + 12²

⇒ r² = 576 + r² - 48r + 144

⇒ r² - r² + 48r = 720

⇒ 48r = 720

⇒ r = $\dfrac{720}{48}$ = 15 cm.

Hence, radius = 15 cm.

An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.

Answer

From figure,

OA = radius = r cm.

BD = DC = $\dfrac{6}{2}$ = 3 cm (As perpendicular to a chord from the center of the circle bisects it)

In right angle triangle ABD,

⇒ AB² = AD² + BD²

⇒ 6² = AD² + 3²

⇒ AD² = 36 - 9

⇒ AD² = 27

⇒ AD = $\sqrt{27}$ = $3\sqrt{3}$ cm.

OD = AD - AO = $(3\sqrt{3} - r)$ cm.

In right angle triangle OBD,

⇒ OB = radius = r cm.

⇒ OB² = OD² + BD²

⇒ r² = $(3\sqrt{3} - r)$² + 3²

⇒ r² = 27 + r² - $6\sqrt{3}$r + 9

⇒ r² - r² + $6\sqrt{3}$r = 36

⇒ $6\sqrt{3}$r = 36

⇒ r = $\dfrac{36}{6\sqrt{3}} = 2\sqrt{3}$ cm.

Hence, radius = $2\sqrt{3}$ cm.

AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.

Answer

Given,

AM = 18 cm and MB = 8 cm

From figure,

AB = AM + MB = 18 + 8 = 26 cm

Radius of the circle = $\dfrac{26}{2}$ = 13 cm.

Let CD is the shortest chord drawn through M.

∴ CD ⊥ AB

From figure,

OM = AM - AO = 18 - 13 = 5 cm

OC (radius) = OA = 13 cm

Now in right ∆OMC,

OC² = OM² + MC² (By pythagoras theorem)

13² = 5² + MC²

MC² = 13² - 5²

MC² = 169 - 25 = 144

MC = $\sqrt{144}$ = 12 cm.

M is Mid-Point of CD (As perpendicular from center to the chord bisects it.)

CD = 2 × MC = 2 × 12 = 24 cm.

Hence, length of shortest chord = 24 cm.

A rectangle with one side of length 4 cm is inscribed in a circle of diameter 5 cm. Find the area of rectangle.

Answer

Let side BC = 4 cm.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ BM = MC = 2 cm.

Given,

Diameter = 5 cm, radius = $\dfrac{5}{2}$ = 2.5 cm.

In right ∆OBM,

⇒ OB² = BM² + OM²

⇒ (2.5)² = 2² + OM²

⇒ 6.25 = 4 + OM²

⇒ OM² = 2.25

⇒ OM = $\sqrt{2.25}$ = 1.5 cm.

Similarly in right ∆OAN,

⇒ OA² = AN² + ON²

⇒ (2.5)² = 2² + ON²

⇒ 6.25 = 4 + ON²

⇒ ON² = 2.25

⇒ ON = $\sqrt{2.25}$ = 1.5 cm.

From figure,

⇒ MN = OM + ON = 1.5 + 1.5 = 3 cm.

⇒ AB = DC = MN = 3 cm.

⇒ Area = length × breadth = 4 cm × 3 cm = 12 cm².

Hence area of rectangle = 12 cm².

The length of the common chord of two intersecting circles is 30 cm. If the radii of the two circles are 25 cm and 17 cm, find the distance between their centres.

Answer

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AC = CB = $\dfrac{30}{2}$ = 15 cm.

From figure,

In right triangle OAC,

⇒ OA² = OC² + AC² (By pythagoras theorem)

⇒ 25² = OC² + 15²

⇒ 625 = OC² + 225

⇒ OC² = 400

⇒ OC = $\sqrt{400}$ = 20 cm.

In right triangle O'AC,

⇒ O'A² = O'C² + AC² (By pythagoras theorem)

⇒ 17² = O'C² + 15²

⇒ 289 = O'C² + 225

⇒ O'C² = 64

⇒ O'C = $\sqrt{64}$ = 8 cm.

Distance between centers = OO' = OC + O'C = 20 + 8 = 28 cm.

Hence, distance between their centres = 28 cm.

The line joining mid-points of two chords of a circle passes through its center. Prove that the chords are parallel.

Answer

In the figure, AB and CD are the two chords of a circle with center O. M and N are mid-points of AB and CD, respectively and MN is the line joining the mid-points of two chords and passing through center O.

Since, the straight line drawn from the centre of a circle to bisect a chord is perpendicular to the chord,

∴ OM ⊥ AB and ON ⊥ CD.

So,

∠OMA = ∠OMB = 90° and ∠ONC = ∠OND = 90°

Since, ∠OMA = ∠OND = 90° (Alternate angles) and,

∠OMB = ∠ONC = 90° (Alternate angles)

Hence, proved that AB || CD.

If a diameter of a circle is perpendicular to one of two parallel chords of the circle, prove that it is perpendicular to the other and bisects it.

Answer

Since, AB || CD and ∠OMA = ∠OMB = 90°

From figure,

∠OMA = ∠OND = 90° (Alternate angles are equal)

∠OMB = ∠ONC = 90° (Alternate angles are equal)

∴ ON ⊥ CD or MN ⊥ CD

We know that,

The perpendicular to a chord from the center of the circle bisects the chord.

∴ NC = ND.

Hence, proved that diameter is perpendicular to other chord and bisects it.

In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.

Answer

From figure,

AD, BE and CF are medians of the triangle.

Let G be the centroid of triangle ABC.

Triangle ABC is an equilateral triangle,

∴ AB = BC = CA and ∠ABC = ∠BAC = ∠BCA = 60°

In △BFC and △BEC,

⇒ BC = BC (Common Side)

⇒ ∠FBC = ∠ECB = 60°.

⇒ BF = EC (As F is mid-point of AB and E is mid-point of AC and AB = AC.)

△BFC ≅ △BEC (By SAS axiom.)

∴ BE = CF (By C.P.C.T.) .........(1)

Now, in △ABE and △ABD

AB = AB (Common Side)

∠BAE = ∠ABD = 60°

BD = AE (As D is mid-point of BC and E is mid-point of AC and BC = AC.)

△ABE ≅ △ABD (By SAS axiom.)

∴ BE = AD (By C.P.C.T.) ............. (2)

From equation 1 and 2, we get:

⇒ AD = BE = CF

⇒ $\dfrac{2}{3}AD = \dfrac{2}{3}BE = \dfrac{2}{3}CF$

We know that G (the centroid) of the triangle divides the median in a 2 : 1 ratio.

∴ GA = GB = GC.

So, we can say that G is equidistant from the three vertices A. B and C.

G is circumcentre of ΔABC.

Hence, proved that the centroid and circumcentre are coincident.

In the figure (i) given below, OD is perpendicular to the chord AB of a circle whose center is O. If BC is a diameter, show that CA = 2OD.

Answer

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AD = DB

We can say that D is mid-point of AB.

Since, BC is diameter and O is center so, OB = OC = radius.

We can say that O is mid-point of BC.

In △ABC,

Since, D is mid-point of AB and O is mid-point of BC

By mid-point theorem,

⇒ OD || AC and OD = $\dfrac{1}{2}AC$

⇒ AC = 2OD.

Hence, proved that CA = 2OD.

In the figure (ii) given below, O is the center of a circle. If AB and AC are chords of the circle such that AB = AC and OP ⊥ AB, OQ ⊥ AC, prove that PB = QC.

Answer

Let AB = AC = x

Given,

OM ⊥ AB and ON ⊥ AC

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ AM = MB = $\dfrac{x}{2}$

and

AN = NC = $\dfrac{x}{2}$

∴ MB = NC ..........(1)

Since, equal chords of a circle are equidistant from the centre,

∴ ON = OM = y (let).

Let radius of circle be r.

From figure,

OQ = OP = r

QN = OQ - ON = r - y

PM = OP - OM = r - y

∴ QN = PM ..........(2)

In △QNC and △PMB,

NC = MB [From (1)]

QN = PM [From (2)]

∠QNC = ∠PMB (Both equal to 90°)

△QNC ≅ △PMB by SAS axiom.

∴ PB = QC (By C.P.C.T.)

Hence, proved that PB = QC.

In the figure (i) given below, a line l intersects two concentric circles at the points A, B, C and D. Prove that AB = CD.

Answer

Draw OM perpendicular to BC.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

So, in the smaller circle, M is mid-point of BC so,

BM = MC = x (let)

Similarly, in larger circle M is mid-point of AD so,

AM = MD = y (let)

From figure,

AB = AM - BM = (y - x)

CD = MD - MC = (y - x)

∴ AB = CD.

Hence, proved that AB = CD.

In the figure (ii) given below, chords AB and CD of a circle with centre O intersect at E. If OE bisects ∠AED, prove that AB = CD.

Answer

Draw perpendiculars from O to AB and CD.

In △OME and △ONE,

∠OME = ∠ONE = 90°

∠OEM = ∠OEN (As OE bisects ∠AED)

OE = OE (Common side)

∴ △OME ≅ △ONE (By A.A.S. axiom)

∴ OM = ON (By C.P.C.T.C.)

Hence, chords AB and CD are equidistant from the center of circle.

In the same circle, chords equidistant from the centre are equal.

∴ AB = CD.

Hence, proved that AB = CD.

In the figure (i) given below, AD is a diameter of a circle with center O. If AB || CD, prove that AB = CD.

Answer

Draw OM ⊥ AB and ON ⊥ CD,

In △OAM and △ODN,

OA = OD (Radius of circle)

∠AOM = ∠DON (Vertically opposite angles are equal)

∠OMA = ∠OND (Both equal to 90°)

∴ △OAM ≅ △ODN (By A.S.A. axiom).

∴ ND = AM (By C.P.C.T.) ........(1)

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

so, equation 1 can be written as,

⇒ $\dfrac{\text{CD}}{2} = \dfrac{\text{AB}}{2}$

⇒ AB = CD.

Hence, proved that AB = CD.

In the figure (ii) given below, AB and CD are equal chords of a circle with center O. If AB and CD meet at E (outside the circle) prove that

(i) AE = CE

(ii) BE = DE.

Answer

Draw ON ⊥ CD and OM ⊥ AB. Join OE.

(i) Since, equal chords are equidistant from the center of the circle,

∴ ON = OM.

In △ONE and △OME,

ON = OM

∠ONE = ∠OME (Both equal to 90°)

OE = OE (Common side)

∴ △ONE ≅ △OME (By R.H.S. congruence rule).

∴ NE = ME = y (let) (By C.P.C.T.) ........(1)

Let AB = CD = x.

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ CN = ND = $\dfrac{CD}{2} = \dfrac{x}{2}$ and

AM = MB = $\dfrac{AB}{2} = \dfrac{x}{2}$.

From figure,

AE = AM + ME = $\dfrac{x}{2} + y$

CE = CN + NE = $\dfrac{x}{2} + y$

Hence, proved that AE = CE.

(ii) From figure,

BE = ME - MB = $y - \dfrac{x}{2}$

DE = NE - ND = $y - \dfrac{x}{2}$

Hence, proved that BE = DE.